Rd Sharma 2022 for Class 9 Maths Chapter 5 - Factorization Of Algebraic Expressions

Share with your friends

Rd Sharma 2022 Solutions for Class 9 Maths Chapter 5 Factorization Of Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Factorization Of Algebraic Expressions are extremely popular among Class 9 students for Maths Factorization Of Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2022 Book of Class 9 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2022 Solutions. All Rd Sharma 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 107:

Question 1:

Factorize:

1. $x^{3} + x - 3 x^{2} - 3$

Answer:

The given expression to be factorized is

Take common x from the first two terms and -3 from the last two terms. That is

Finally, take common x² + 1from the two terms. That is

We cannot further factorize the expression.

So, the required factorization is.

Page No 107:

Question 2:

Factorize:

2. a(a+b)³ − 3a²b (a + b)

Answer:

The given expression to be factorized is

Take common from the two terms. That is

Expand the term within the second braces.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 3:

$x (x^{3} - y^{3}) + 3 x y (x - y)$

Answer:

The given expression to be factorized is

We know that

The given expression then becomes

Take common from the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 4:

Factorize:

a²x² + (ax² + 1)x + a

Answer:

The given expression to be factorized is

Simplify the middle term. That is

Take common from the first two terms and 1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 5:

Factorize:
x² + y − xy − x

Answer:

The given expression to be factorized is

Rearrange the given expression as

Take common x from the first two terms and -1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 6:

Factorize:

x³ − 2x²y + 3xy² − 6y³

Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 7:

Factorize:

6ab − b² + 12ac − 2bc

Answer:

The given expression to be factorized is

Take common b from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 8:

Factorize:

$x (x - 2) (x - 4) + 4 x - 8$

Answer:

The given expression to be factorized is

Take common 4 from the last two terms. That is

Again take commonfrom the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 9:

Factorize:

$21 x^{2} - 2 x + \frac{1}{21}$

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 10:

Factorize:

a² + 2ab +b² − c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 11:

Factorize:

a² + 4b² − 4ab − 4c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitute.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 12:

Factorize:

x² − y² − 4xz + 4z²

Answer:

The given expression to be factorized is

Rearrange the terms as

Substitutingin the avove expression,

Put.

We cannot further factorize the expression.

So, the required factorization ofis.

Page No 107:

Question 13:

Factorize:

$2 x^{2} - \frac{5}{6} x + \frac{1}{12}$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 14:

Factorize:

$x^{2} + \frac{12}{35} x + \frac{1}{35}$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 15:

Factorize:

(a − b + c)² + (b − c + a)² + 2(a − b + c) (b − c + a)

Answer:

The given expression to be factorized is

This can be written as

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 16:

Give possible expressions for the length and breadth of the rectangle having 35y² + 13y − 12 as its area.

Answer:

The area of the rectangle is

First we will factorize the above expression. This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

The area of a rectangle having length a and breadth bis ab.

Here we don’t know the bigger or the smaller factor. So, the two possibilities are

(i) Length isand breadth is

(ii) Length is and breadth is

Page No 107:

Question 17:

What are the possible expressions for the dimensions of the cuboid whose volume is 3x²− 12x.

Answer:

The volume of the cuboid is

First we will factorize the above expression.

Take commonfrom the two terms of the above expression,

The volume of a cuboid having length, breadth b and height is.

Here the word ‘dimensions’ stands for the length, breadth and height of the cuboid. So, the three possibilities are

(i) Length is, breadth is x and height is

(ii) Length is x , breadth isand height is

(iii) Length is, breadth isand height is x

There are many other possibilities also, because we can consider the product of two simple factors as a single factor.

Page No 107:

Question 18:

Factorize:
a² + b² + 2bc − c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 19:

Factorize:

xy⁹ − yx⁹

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the two terms of the above expression

We cannot further factorize the expression.

So, the required factorization of is

Page No 107:

Question 20:

Factorize:

x⁴+ x²y² + y⁴

Answer:

The given expression to be factorized is

Add and subtract the termin the given expression.

Substitutingin the above expression, we get

Putin the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 21:

Factorize:

$x^{2} + 6 \sqrt{2} x + 10$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms.

Finally, take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization is.

Page No 107:

Question 22:

Factorize:

$x^{2} - 2 \sqrt{2} x - 30$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 23:

Factorize:

$x^{2} - \sqrt{3} x - 6$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 24:

Factorize:

$x^{2} + 5 \sqrt{5} x + 30$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 25:

Factorize:

$x^{2} + 2 \sqrt{3} x - 24$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 26:

Factorize:

$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 27:

Factorize:

$2 x^{2} + 3 \sqrt{5} x + 5$

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 28:

Factorize:

$2 a^{2} + 2 \sqrt{6} a b + 3 b^{2}$

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 29:

Factorize:

7(x − 2y)² − 25(x − 2y) + 12

Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put in the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 30:

Factorize:

2(x + y)² − 9(x + y) − 5

Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 31:

Factorize:

4(x − y)² − 12(x − y) (x + y) + 9(x + y)²

Answer:

The given expression to be factorized is

Substituting andin the above expression, we get

This can be arrange in the form

Putand.

Take common -1 from the expression within the braces.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 32:

Factorize:

a² + b² + 2(ab + bc + ca)

Answer:

The given expression to be factorized is

This can be written as

Take commonfrom the last two terms.

Finally, take common from the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 33:

Factorize:

(x+2) (x²+25) − 10x² − 20x

Answer:

The given expression to be factorized is

Take common from the last two terms. That is

Again take commonfrom the two terms of the above expression. Then

We cannot further factorize the expression.

So, the required factorization of is.

Page No 107:

Question 34:

Factorize:

$(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$

Answer:

The given expression to be factorized is

We have

Use the above result in the original expression to get

Substituting in the above , we get

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 1:

y³ + 125

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 2:

1 − 27a³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 3:

8x³y³ + 27a³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 4:

64a³ − b³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 5:

$\frac{x^{3}}{216} - 8 y^{3}$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 6:

10x⁴y − 10xy⁴

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 7:

54x⁶y + 2x³y⁴

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 8:

32a³ + 108b³

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 9:

(a − 2b)³ − 512b³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 10:

8x²y³ − x⁵

Answer:

The given expression to be factorized is

Take common. Then we have

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 11:

1029 − 3x³

Answer:

The given expression to be factorized is

Take common 3. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 12:

x³y³+ 1

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 13:

x⁴y⁴ − xy

Answer:

The given expression to be factorized is

Take common. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 14:

a³ + b³ + a + b

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

Take common. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 15:

Simplify:

(i) $\frac{173 \times 173 \times 173 + 127 \times 127 \times 127}{173 \times 173 - 173 \times 127 + 127 \times 127}$

(ii) $\frac{155 \times 155 \times 155 - 55 \times 55 \times 55}{155 \times 155 + 155 \times 55 + 55 \times 55}$

(iii) $\frac{1.2 \times 1.2 \times 1.2 - 0.2 \times 0.2 \times 0.2}{1.2 \times 1.2 + 1.2 \times 0.2 + 0.2 \times 0.2}$

Answer:

(i) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and b are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(ii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, b is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(iii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, b is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

Page No 111:

Question 16:

(a + b)³ − 8(a − b)³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 17:

(x + 2)³ + (x − 2)³

Answer:

The given expression to be factorized is

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 18:

a¹²+ b¹²

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 19:

x³ + 6x² + 12x + 16

Answer:

The given expression to be factorized is

This can be written as

Take common x² from first two terms, 2x from the next two terms andfrom the last two terms. Then we have,

Finally, take common. Then we get,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 20:

$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 21:

a³ + 3a²b + 3ab² + b³ − 8

Answer:

The given expression to be factorized is

Recall the well known formula

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common –2 from the last two terms, we get

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 22:

8a³ − b³ − 4ax + 2bx

Answer:

The given expression to be factorized is

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 111:

Question 23:

Simplify: (2x – 5y)³ – (2x + 5y)³

Answer:

Ans

Page No 114:

Question 1:

Factorize:

64a³ + 125b³ + 240a²b + 300ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 114:

Question 2:

125x³ − 27y³ − 225x²y + 135xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 3:

$\frac{8}{27} x^{3} + 1 + \frac{4}{3} x^{2} + 2 x$

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 4:

8x³ + 27y³ + 36x²y + 54xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 5:

a³ − 3a²b + 3ab² − b³ + 8

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the third and fourth terms. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

This can be written in the following form

Recall the formula for the sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 114:

Question 6:

x³ + 8y³ + 6x²y + 12xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 7:

8x³ + y³ + 12x²y + 6xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common 6xy from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 8:

8a³+ 27b³ + 36a²b + 54ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 9:

8a³ − 27b³ − 36a²b + 54ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common – 18ab from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 114:

Question 10:

x³ − 12x(x − 4) − 64

Answer:

The given expression to be factorized is

This can be written in the form

Take common – 12x from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 114:

Question 11:

a³x³ − 3a²bx² + 3ab²x − b³

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 119:

Question 1:

Factorize each of the following expressions:

a³ + 8b³ + 64c³ − 24abc

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 119:

Question 2:

x³− 8y³ + 27z³ + 18xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 119:

Question 3:

27x³ − y³ − z³ − 9xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 119:

Question 4:

$\frac{1}{27} x^{3} - y^{3} + 125 z^{3} + 5 x y z$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 119:

Question 5:

8x³ +27y³ − 216z³ + 108xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 119:

Question 6:

125 + 8x³ − 27y³ + 90xy

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 119:

Question 7:

8x³ − 125y³ + 180xy + 216

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 119:

Question 8:

Multiply:
(i) x² + y² + z² − xy + xz + yz by x + y − z

(ii) x² + 4y² + z² + 2xy + xz − 2yz by x − 2y − z

(iii) x²+ 4y² + 2xy − 3x + 6y + 9 by x − 2y + 3

(iv) 9x² + 25y² + 15xy + 12x − 20y + 16 by 3x − 5y + 4

(v) x²+ 4y² + z² + 2xy + xz – 2yz by (−z + x – 2y)

(vi) 4x² + y² + 9z² + 2xy + 3yz – 6xz by (2x – y + 3z)

Answer:

(i) The given expression is .

We have to multiply the above expression by.

The required product is

Recall the formula

Using the above formula, we have

(ii) The given expression is .

We have to multiply the above expression by.

The required product is

Recall the formula

Using the above formula, we have

(iii) The given expression is .

We have to multiply the above expression by.

The required product is

Recall the formula

Using the above formula, we have

(iv) The given expression is .

We have to multiply the above expression by.

The required product is

Recall the formula

Using the above formula, we have

(v) The given expression is

We have to multiply the above expression by $(- z + x - 2 y)$ .

The required product is

Recall the formula

Using the above formula, we have

(vi) The given expression is 4x² + y² + 9z² + 2xy + 3yz – 6xz.

We have to multiply the above expression by (2x – y + 3z).

The required product is

(2x – y + 3z)(4x² + y² + 9z² + 2xy + 3yz – 6xz)
= {2x + (– y) + 3z}{(2x)² + (– y)² + (3z)² – (2x)(–y) – (–y)(3z) – (3z)(2x)}

Recall the formula

Using the above formula, we have

= (2x)³ + (–y)³ + (3z)³ – 3(2x)(–y)(3z)
= 8x³ – y³ + 27z³ + 18xyz

Page No 119:

Question 9:

$2 {\sqrt{2 a}}^{3} + 3 {\sqrt{3 b}}^{3} + c^{3} - 3 \sqrt{6} a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 119:

Question 10:

$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 119:

Question 11:

$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 119:

Question 12:

Without actually computing the cubes, find the value of 48³ – 30³ – 18³.

Answer:

$48^{3} - 30^{3} - 18^{3} = 48^{3} + {(- 30)}^{3} + {(- 18)}^{3}$
Let x = 48, y = −30, z = −18
$∴ x + y + z = 0$
We know,
if x + y + z = 0
then, x³ + y³ + z³ = 3xyz

$\begin{array}{rcl} ∴ 48^{3} + {(- 30)}^{3} + {(- 18)}^{3} & = & 3 (48) (- 30) (- 18) \\ = & 77760 \end{array}$

Page No 119:

Question 13:

(3x − 2y)³ + (2y − 4z)³ + (4z − 3x)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 119:

Question 14:

(2x − 3y)³ + (4z − 2x)³ + (3y − 4z)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 119:

Question 15:

(a − 3b)³ + (3b − c)³ + (c − a)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 119:

Question 16:

(x – 2y)³ + (2y – 3z)³ + (3z – x)³

Answer:

$To solve : {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} Let a = (x - 2 y), b = (2 y - 3 z), c = (3 z - x) . a + b + c = (x - 2 y) + (2 y - 3 z) + (3 z - x) = x - 2 y + 2 y - 3 z + 3 z - x = 0 As we know that, if a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3 a b c Therefore, a^{3} + b^{3} + c^{3} = 3 a b c \Rightarrow {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x) Hence, {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

Page No 119:

Question 17:

${(\frac{x}{2} + y + \frac{z}{3})}^{3} + {(\frac{x}{3} - \frac{2 y}{3} + z)}^{3} + {(- \frac{5 x}{6} - \frac{y}{3} - \frac{4 z}{3})}^{3}$

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and.

Then we have

We cannot further factorize the expression.

So, the required factorization is of is

Page No 119:

Question 18:

Find the value of x³ + y³ − 12xy + 64, when x + y =−4

Answer:

The given expression is

It is given that

The given expression can be written in the form

Recall the formula

Using the above formula, we have

Page No 119:

Question 19:

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3$ .

Answer:

$Given : a + b + c = 0 As we know that, if a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3 a b c Therefore, a^{3} + b^{3} + c^{3} = 3 a b c Dividing by (a b c) on both sides, we get \Rightarrow \frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{3 a b c}{a b c} \Rightarrow \frac{a^{3}}{a b c} + \frac{b^{3}}{a b c} + \frac{c^{3}}{a b c} = 3 \Rightarrow \frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = 3 Hence, \frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = 3 .$

Page No 120:

Question 1:

If a + b + c = 0, then write the value of a³ + b³ + c³.

Answer:

Recall the formula

When, we have

Page No 120:

Question 2:

If a² + b² + c² = 20 and a + b + c = 0, find ab + bc + ca.

Answer:

Recall the formula

Given that

Then we have

Page No 120:

Question 3:

If a + b + c = 9 and ab + bc + ca = 40, find a² + b² +c².

Answer:

Recall the formula

Given that

Then we have

Page No 120:

Question 4:

If a² + b² + c² = 250 and ab + bc + ca = 3, find a + b + c.

Answer:

Recall the formula

Given that

Then we have

Page No 120:

Question 5:

Write the value of 25³ − 75³+ 50³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 120:

Question 6:

Write the value of 48³ − 30³ − 18³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 120:

Question 7:

Write the value of ${(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3} .$

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 120:

Question 8:

Write the value of 30³ + 20³ − 50³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 120:

Question 9:

Factorize: x⁴ + x² + 25.

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization is.

Page No 120:

Question 10:

Factorize : x² − 1 − 2a − a²

Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

We cannot further factorize the expression.

So, the required factorization is.

Page No 121:

Question 1:

The factorized form of the expression y² + (x – 1)y – x is ____________.

Answer:

$y^{2} + (x - 1) y - x = y^{2} + x y - y - x = y (y + x) - 1 (y + x) = (y - 1) (y + x)$

Hence, the factorized form of the expression y² + (x – 1)y – x is (y – 1)(y + x).

Page No 121:

Question 2:

The factorized form of a³ + (b – a)³ – b³ is ____________.

Answer:

$a^{3} + {(b - a)}^{3} - b^{3} = a^{3} + (b^{3} - a^{3} - 3 b a (b - a)) - b^{3} (Using the identity : {(x - y)}^{3} = x^{3} - y^{3} - 3 x y (x - y)) = a^{3} + b^{3} - a^{3} - 3 b a (b - a) - b^{3} = - 3 b a (b - a) = 3 a b (a - b)$

Hence, the factorized form of a³ + (b – a)³ – b³ is 3ab(a – b).

Page No 121:

Question 3:

If $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c =____________.$

Answer:

$Given : \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 . . . (1) a b c = 2 . . . (2) Now, a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = a^{2} b^{2} c^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(a b c)}^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(2)}^{2} (1) (From (1) and (2)) = 4$

Hence, if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = 4 .$

Page No 121:

Question 4:

Factorization of the polynomial $11 x^{2} - 10 \sqrt{3} x - 3$ gives ____________.

Answer:

$11 x^{2} - 10 \sqrt{3} x - 3 = 11 x^{2} - 11 \sqrt{3} x + \sqrt{3} x - 3 = 11 x (x - \sqrt{3}) + \sqrt{3} (x - \sqrt{3}) = (11 x + \sqrt{3}) (x - \sqrt{3})$

Hence, factorization of the polynomial $11 x^{2} - 10 \sqrt{3} x - 3$ gives $(11 x + \sqrt{3}) (x - \sqrt{3})$ .

Page No 121:

Question 5:

The polynomial x² + y² – z² – 2xy on factorization gives _____________.

Answer:

$x^{2} + y^{2} - z^{2} - 2 x y = (x^{2} + y^{2} - 2 x y) - z^{2} = {(x - y)}^{2} - {(z)}^{2} (Using the identity : {(a - b)}^{2} = a^{2} + b^{2} - 2 a b) = (x - y + z) (x - y - z) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the polynomial x² + y² – z² – 2xy on factorization gives $(x - y + z) (x - y - z)$ .

Page No 121:

Question 6:

The factors of the expression $a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are ____________.

Answer:

$a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a} \sqrt{b} + 2 \sqrt{b} (- \sqrt{c}) + 2 \sqrt{a} (- \sqrt{c}) = {(\sqrt{a} + \sqrt{b} - \sqrt{c})}^{2} (Using the identity : a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c = {(a + b + c)}^{2}) = (\sqrt{a} + \sqrt{b} - \sqrt{c}) (\sqrt{a} + \sqrt{b} - \sqrt{c})$

Hence, the factors of the expression $a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are $(\sqrt{a} + \sqrt{b} - \sqrt{c}) and (\sqrt{a} + \sqrt{b} - \sqrt{c}) .$ .

Page No 121:

Question 7:

The polynomial , x⁶ + 64y⁶ on factorization gives _____________.

Answer:

$x^{6} + 64 y^{6} = {(x^{2})}^{3} + {(4 y^{2})}^{3} = (x^{2} + 4 y^{2}) ({(x^{2})}^{2} + {(4 y^{2})}^{2} - (x^{2}) (4 y^{2})) (Using the identity : a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)) = (x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$

Hence, the polynomial x⁶ + 64y⁶ on factorization gives $(x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$ .

Page No 121:

Question 8:

The factorization form of a⁴ + b⁴– a²b² is _____________.

Answer:

$a^{4} + b^{4} - a^{2} b^{2} Adding and subtracting 2 a^{2} b^{2}, we get = {(a^{2})}^{2} + {(b^{2})}^{2} + 2 a^{2} b^{2} - 2 a^{2} b^{2} - a^{2} b^{2} = {(a^{2} + b^{2})}^{2} - 3 a^{2} b^{2} (Using the identity : a^{2} + b^{2} + 2 a b = {(a + b)}^{2}) = {(a^{2} + b^{2})}^{2} - {(\sqrt{3} a b)}^{2} = (a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the factorization form of a⁴ + b⁴– a²b² is $(a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b)$ .

Page No 121:

Question 9:

If $3 x - \frac{y}{5} = 10 and x y = 5$ , then the value of $27 x^{3} - \frac{y^{3}}{125}$ is ____________.

Answer:

$Given : 3 x - \frac{y}{5} = 10 . . . (1) x y = 5 . . . (2) Now, 3 x - \frac{y}{5} = 10 Taking cube on both sides, we get \Rightarrow {(3 x - \frac{y}{5})}^{3} = 10^{3} \Rightarrow {(3 x)}^{3} - {(\frac{y}{5})}^{3} - 3 (3 x) (\frac{y}{5}) (3 x - \frac{y}{5}) = 1000 (Using the identity : {(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a - b)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} x y (3 x - \frac{y}{5}) = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} \times 5 \times 10 = 1000 (From (1) and (2)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - 90 = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1000 + 90 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1090$

Hence, the value of $27 x^{3} - \frac{y^{3}}{125}$ is 1090.

Page No 121:

Question 10:

The factorized form of $\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is _____________.

Answer:

$\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \frac{(x^{2} + y^{2} + z^{2})}{x y z} + 2 (\frac{y z + x z + x y}{x y x}) = \frac{x^{2} + y^{2} + z^{2} + 2 (y z + x z + x y)}{x y z} = \frac{{(x + y + z)}^{2}}{x y z} (Using the identity : {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c) = \frac{1}{x y z} {(x + y + z)}^{2}$

Hence, the factorized form of $\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is $\frac{1}{x y z} {(x + y + z)}^{2} .$

Page No 121:

Question 11:

The factorized form of a³ + b³ + 3ab – 1 is ____________.

Answer:

$a^{3} + b^{3} + 3 a b - 1 = a^{3} + b^{3} - 1 + 3 a b = a^{3} + b^{3} + {(- 1)}^{3} - 3 a b (- 1) = (a + b - 1) (a^{2} + b^{2} + {(- 1)}^{2} - a b - b (- 1) - a (- 1)) (Using the identity : a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)) = (a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a)$

Hence, the factorized form of a³ + b³ + 3ab – 1 is $(a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a) .$

View NCERT Solutions for all chapters of Class 9