R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 4 - Algebraic Identities

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Hence the value of is

Hence the correct choice is (c).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (d).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Squaring on both sides we get, 

Hence the value of is

Hence the correct choice is (b).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identityand

Here put,

Take Cube on both sides we get,

Hence the value of is

Hence the correct choice is (d).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (b).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Put we get,

Substitute y = 5 in the above equation we get

The Equation satisfy the condition that

Hence the value of is 5

The correct choice is (a).

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Put we get,

Substitute y = 2 in above equation we get,

The Equation satisfy the condition that

Hence the value of is 2

Hence the correct choice is (d).

Answer:

We have to find

Given

Using identity we get,

By transposing +46 to left hand side we get,

Hence the value of is

The correct choice is (a).

Answer:

Given

Using identity

Here

Hence the Value of is

The correct choice is .

Answer:

Given: a + b = 3  .....(1)
And ab = 2          .....(2)

$$\begin{gathered} \mathrm{Cubing} \left(1\right) \mathrm{on} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\left(a+b\right)}^3=3^3 \\ \Rightarrow a^3+b^3+3ab\left(a+b\right)=27 \\ \Rightarrow a^3+b^3+3\left(2\right)\left(3\right)=27 \left[\mathrm{From}\left(1\right) \mathrm{and}\left(2\right)\right] \\ \Rightarrow a^3+b^3+18=27 \\ \Rightarrow a^3+b^3=27-18 \\ \Rightarrow a^3+b^3=9 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

To find the value of a³ − b³ 

Given

Using identity

Here we get

Transposing -288 to left hand side we get 

Hence the value of is -224

The correct choice is .

Answer:

We have to find the possible dimension of cuboid 

Given: volume of cuboid 

Take 3 as common factor

Using identity

We get,

Here the dimension of cuboid is 3,

The correct alternate is .

Answer:

We have to find the product of

Using identity

Here

Hence the product of is 10,000

The correct choice is .

Answer:

Given

Using the identity

Hence is equal to

The correct choice is .

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

We shall use the identitywe get,

Taking square root on both sides we get,

Hence the value of is

Hence the correct choice is (c).

Answer:

Given

Using identity

Here,

Again using identity

Here

Substituting

Using identity

Here

Hence the value of is

The correct choice is (b).  

Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Substitute in we get,

Hence the value of is

Hence the correct choice is (b).

Answer:

We have to find the value of

Given

Using identity we get,

Here

Substituting we get,

By transposing left hand side we get,

Again using identity we get,

Substituting we get 

Using identity we get 

Here

Substituting we get,

The value of is

The correct choice is (b) 

Answer:

Given

Multiplying both sides by 2 we get,

Therefore the sum of positive quantities is zero if and only if each quantity is zero.

If, then

The correct choice is (d).

Answer:

We have to find

Given

Using identity

Hence the value of

The correct choice is (d).

Answer:

Given

Using identity we get

Here

Taking Cube on both sides we get,

Hence the value of is

The correct choice is .

Answer:

We have to find the value of

Given

Using identity we get,

By transposing +46 to left hand side we get,

Using identity

The value of is

Hence the correct choice is .

Answer:

We have to find the value of

Using Identity we get,

Hence the value of is

The correct choice is .

Answer:

We have to find the product of

Using identity 

We can rearrange as 

Again using the identity

Here

Hence the product of is

The correct choice is .

Answer:

We have to find the product of

Using identity

Here

Hence the product value of is

The correct alternate is .

Answer:

Given

Using identity we get,

Hence the value of is .

The correct choice is (d).

Answer:

We have to find the value of b

Given

Using identity

We get

Equating ‘b’ on both sides we get 

Hence the value of b is

The correct choice is .

Answer:

$$\begin{gathered} \left(25x^2-1\right)+{\left(1+5x\right)}^2 \\ =\left({\left(5x\right)}^2-{\left(1\right)}^2\right)+{\left(1+5x\right)}^2 \\ =\left(5x-1\right)\left(5x+1\right)+\left(1+5x\right)\left(1+5x\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ =\left(5x+1\right)\left(5x-1+1+5x\right) \\ =\left(5x+1\right)\left(10x\right) \\ \mathrm{Therefore}, \left(25x^2-1\right)+{\left(1+5x\right)}^2 \mathrm{has} \mathrm{two} \mathrm{factors} \left(5x+1\right) \mathrm{and} \left(10x\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ \Rightarrow 9x^2-b={\left(3x\right)}^2-{\left(\frac{1}{2}\right)}^2 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ \Rightarrow 9x^2-b=9x^2-\frac{1}{4} \\ \Rightarrow -b=-\frac{1}{4} \\ \Rightarrow b=\frac{1}{4} \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right). \end{gathered}$$

Answer:

$$\begin{gathered} {\left(x+3\right)}^3 \\ ={\left(x\right)}^3+{\left(3\right)}^3+3\left(x\right)\left(3\right)\left(x+3\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ =x^3+27+9x\left(x+3\right) \\ =x^3+27+9x^2+27x \\ =x^3+9x^2+27x+27 \\ \mathrm{Thus}, \mathrm{the} \mathrm{coefficient} \mathrm{of} x \mathrm{is} 27. \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Answer:

$$\begin{gathered} {\left(249\right)}^2-{\left(248\right)}^2 \\ =\left(249+248\right)\left(249-248\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ =\left(497\right)\left(1\right) \\ =497 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Answer:

$$\begin{gathered} {\left(x+y\right)}^3-\left(x^3+y^3\right) \\ ={\left(x\right)}^3+{\left(y\right)}^3+3\left(x\right)\left(y\right)\left(x+y\right)-\left(x^3+y^3\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ =x^3+y^3+3xy\left(x+y\right)-x^3-y^3 \\ =3xy\left(x+y\right) \\ \mathrm{Thus}, {\left(x+y\right)}^3-\left(x^3+y^3\right) \mathrm{has} \mathrm{two} \mathrm{factors} \left(3xy\right) \mathrm{and} \left(x+y\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ \frac{x}{y}+\frac{y}{x}=-1 \\ \Rightarrow \frac{x^2+y^2}{xy}=-1 \\ \Rightarrow x^2+y^2=-xy \\ \Rightarrow x^2+y^2+xy=0 ...\left(1\right) \\ \mathrm{Now}, \\ \left(x^3-y^3\right) \\ =\left(x-y\right)\left(x^2+y^2+xy\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right) \\ =\left(x-y\right)\times 0 \left(\mathrm{From} \left(1\right)\right) \\ =0 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right). \end{gathered}$$

Answer:

Given: x + y = 2  .....(1)
And xy = 1          .....(2)

Squaring (1) on both the sides
$$\begin{gathered} {\left(x+y\right)}^2=2^2 \\ \Rightarrow x^2+2xy+y^2=4 \\ \Rightarrow x^2+2\left(1\right)+y^2=4 \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow x^2+y^2=2 .....\left(3\right) \end{gathered}$$

Squaring (3) on both the sides
$$\begin{gathered} {\left(x^2+y^2\right)}^2=2^2 \\ \Rightarrow {\left(x^2\right)}^2+2x^2{y}^2+{\left(y^2\right)}^2=4 \\ \Rightarrow x^4+2{\left(xy\right)}^2+y^4=4 \\ \Rightarrow x^4+2{\left(1\right)}^2+y^4=4 \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow x^4+y^4=2 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

$$\begin{gathered} \mathrm{Given}: \\ {x}^2+y^2+xy=1 ...\left(1\right) \\ x+y=2 ...\left(2\right) \\ \mathrm{Now}, \\ x+y=2 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(x+y\right)}^2=2^2 \\ \Rightarrow x^2+y^2+2xy=4 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow x^2+y^2+xy+xy=4 \\ \Rightarrow 1+xy=4 \left(\mathrm{From} \left(1\right)\right) \\ \Rightarrow xy=4-1 \\ \Rightarrow xy=3 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{b}\right). \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ {a}^2+b^2+c^2=29 ...\left(1\right) \\ ab+bc+ca=26 ...\left(2\right) \\ \mathrm{Now}, \\ {\left(a+b+c\right)}^2 \\ =a^2+b^2+c^2+2\left(ab+bc+ca\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\right) \\ =29+2\left(26\right) \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =29+52 \\ =81 \\ \mathrm{Since}, {\left(a+b+c\right)}^2=81 \mathrm{and} a, b, c \mathrm{are} \mathrm{natural} \mathrm{numbers} \\ \mathrm{Therefore}, a+b+c=9. \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{a}\right). \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ 2x+\frac{y}{3}=12 ...\left(1\right) \\ xy=30 ...\left(2\right) \\ \mathrm{Now}, \\ 2x+\frac{y}{3}=12 \\ \mathrm{Taking} \mathrm{cube} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(2x+\frac{y}{3}\right)}^3={12}^3 \\ \Rightarrow {\left(2x\right)}^3+{\left(\frac{y}{3}\right)}^3+3\left(2x\right)\left(\frac{y}{3}\right)\left(2x+\frac{y}{3}\right)=1728 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+2xy\left(2x+\frac{y}{3}\right)=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}+2\times 30\times 12=1728 \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+720=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1728-720 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1008 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{a}\right). \end{gathered}$$

Answer:

Statement-2 (Reason): a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right)& =& a^3+a{b}^{2 }+a{c}^{2 }–a^{2}b–abc–c{a}^2+b{a}^2+b^{3 }+b{c}^{2 }–a{b}^2–b^{2}c–cab+c{a}^2+c{b}^{2 }+c^{3 }–abc–b{c}^2–c^{2}a\\ & =& a^3+b^3+c^3+a{b}^{2 }-a{b}^2+a{c}^{2 }–c^{2}a+b{a}^2–a^{2}b+c{a}^2–c{a}^2+b{c}^{2 }–b{c}^2+c{b}^2–b^{2}c–cab–abc–abc\\ & =& a^3+b^3+c^3–3abc\end{array}$

Thus, Statement-2 is true.

Statement-1 (Assertion): If a + b + c = 0, then a³ + b³ + c³ = 3abc

Now, According to the Statement-2

$$\begin{gathered} \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0·\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0 \\ \Rightarrow a^3+b^3+c^3=3abc \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).
 

Answer:

Statement-2 (Reason): a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right)& =& a^3+a{b}^{2 }+a{c}^{2 }–a^{2}b–abc–c{a}^2+b{a}^2+b^{3 }+b{c}^{2 }–a{b}^2–b^{2}c–cab+c{a}^2+c{b}^{2 }+c^{3 }–abc–b{c}^2–c^{2}a\\ & =& a^3+b^3+c^3+a{b}^{2 }-a{b}^2+a{c}^{2 }–c^{2}a+b{a}^2–a^{2}b+c{a}^2–c{a}^2+b{c}^{2 }–b{c}^2+c{b}^2–b^{2}c–cab–abc–abc\\ & =& a^3+b^3+c^3–3abc\end{array}$

Thus, Statement-2 is true.

Statement-1 (Assertion): (a + b + c)² = a² + b² + c² – 2(ab + bc + ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a+b+c\right)& =& a^2+ab+ac+ab+b^2+bc+ac+bc+c^2\\ & =& a^2+b^2+c^2+ab+ab+ac+ac+bc+bc\\ & =& a^2+b^2+c^2+2ab+2ac+2bc\\ & =& a^2+b^2+c^2+2\left(ab+ac+bc\right)\end{array}$

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.
 
Hence, the correct answer is option (d).

Answer:

Statement-2 (Reason): a³ + b³ + c³ + 3abc = (a + b + c) (a² + b² + c² + ab + bc + ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }+ab+bc+ca\right)& =& a^3+a{b}^{2 }+a{c}^{2 }+a^{2}b+abc+c{a}^2+b{a}^2+b^{3 }+b{c}^{2 }+a{b}^2+b^{2}c+cab+c{a}^2+c{b}^{2 }+c^{3 }+abc+b{c}^2+c^{2}a\\ & =& a^3+b^3+c^3+a{b}^{2 }+a{b}^2+a{c}^{2 }+a{c}^2+a^{2}b+a^{2}b+c{a}^2+c{a}^2+b{c}^{2 }+b{c}^2+c{b}^2+b^{2}c+cab+abc+abc\\ & =& a^3+b^3+c^3+2a{b}^2+2a{c}^2+2a^{2}b+2a^{2}c+2b^{2}c+2b{c}^2+3abc\end{array}$

Thus, Statement-2 is false.

Statement-1 (Assertion): $a^3+\frac{3}{8}ax+\frac{1}{64}{x}^3-\frac{1}{8}=\left(a+\frac{x}{4}-\frac{1}{2}\right)\left(a^2+\frac{x^2}{16}+\frac{1}{4}-\frac{ax}{4}+\frac{x}{8}+\frac{a}{2}\right)$
Now, using a³ + b³ + c³ − 3abc = (a + b + c) (a² + b² + c² − ab − bc − ca)

$$\begin{gathered} \left(a+\frac{x}{4}-\frac{1}{2}\right)\left(a^2+\frac{x^{2 }}{16}+\frac{1}{4}–\frac{ax}{4}+\frac{x}{8}+\frac{a}{2}\right) \\ =\left(a+\frac{x}{4}-\frac{1}{2}\right)\left[{\left(a\right)}^2+{\left(\frac{x}{4}\right)}^2+{\left(-\frac{1}{2}\right)}^2–\left(a\right)\left(\frac{x}{4}\right)-\left(\frac{x}{4}\right)\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)\left(a\right)\right] \\ =a^3+{\left(\frac{x}{4}\right)}^3+{\left(-\frac{1}{2}\right)}^3-3\left(a\right)\left(\frac{x}{4}\right)\left(-\frac{1}{2}\right) \\ =a^3+\frac{x^3}{64}-\frac{1}{8}+\frac{3ax}{8} \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a+b+c\right)& =& a^2+ab+ac+ab+b^2+bc+ac+bc+c^2\\ & =& a^2+b^2+c^2+ab+ab+ac+ac+bc+bc\\ & =& a^2+b^2+c^2+2ab+2ac+2bc\\ & =& a^2+b^2+c^2+2\left(ab+ac+bc\right)\end{array}$

Thus, Statement-2 is true.

Statement-1 (Assertion): If a + b + c = 6, ab + bc + ca = 11, then a² + b² + c² = 14.

Now, According to the Statement-2.

$$\begin{gathered} a^2+b^2+c^2+2\left(ab+ac+bc\right)={\left(a+b+c\right)}^2 \\ \Rightarrow a^2+b^2+c^2+2\left(11\right)={\left(6\right)}^2 \\ \Rightarrow a^2+b^2+c^2+22=36 \\ \Rightarrow a^2+b^2+c^2=36-22 \\ \Rightarrow a^2+b^2+c^2=14 \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If a + b + c = 0, then a³ + b³ + c³ = 3abc.

$$\begin{gathered} \Rightarrow a^3+b^3+c^3–3abc=\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0·\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right) \\ \Rightarrow a^3+b^3+c^3–3abc=0 \\ \Rightarrow a^3+b^3+c^3=3abc \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): $\frac{{(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}=\left(x+y\right)\left(y+z\right)\left(z+x\right)$

Here, $(x^2-y^2)+(y^2-z^2)+(z^2-x^2)=0$

Now, According to the Statement-2
${(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3=3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)$         .....(1)

Also, $(x-y)+(y-z)+(z-x)=0$

Now, According to the Statement-2
${(x-y)}^3+{(y-z)}^3+{(z-x)}^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)$         .....(2)

Now, 
$\begin{array}{rcl}\frac{{(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}& =& \frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)} \left[\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right]\\ & =& \frac{3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)}{3(x-y)(y-z)(z-x)}\\ & =& \left(x+y\right)\left(y+z\right)\left(z+x\right)\end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

$\begin{array}{rcl}\left(a+b+c\right)\left(a^2+b^{2 }+c^{2 }–ab–bc–ca\right)& =& a^3+a{b}^{2 }+a{c}^{2 }–a^{2}b–abc–c{a}^2+b{a}^2+b^{3 }+b{c}^{2 }–a{b}^2–b^{2}c–cab+c{a}^2+c{b}^{2 }+c^{3 }–abc–b{c}^2–c^{2}a\\ & =& a^3+b^3+c^3+a{b}^{2 }-a{b}^2+a{c}^{2 }–c^{2}a+b{a}^2–a^{2}b+c{a}^2–c{a}^2+b{c}^{2 }–b{c}^2+c{b}^2–b^{2}c–cab–abc–abc\\ & =& a^3+b^3+c^3–3abc\end{array}$

Thus, Statement-2 is true.

Statement-1 (Assertion): The square root of $\frac{1}{abc}(a^2+b^2+c^2)+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is $\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}.$
$\begin{array}{rcl}{\left(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\right)}^2& =& {\left(\sqrt{\frac{a}{bc}}\right)}^2+{\left(\sqrt{\frac{b}{ca}}\right)}^2+{\left(\sqrt{\frac{c}{ab}}\right)}^2+2\left[\left(\sqrt{\frac{a}{bc}\times \frac{b}{ca}}\right)+\left(\sqrt{\frac{b}{ca}\times \frac{c}{ab}}\right)+\left(\sqrt{\frac{c}{ab}\times \frac{a}{bc}}\right)\right] \left[\because {\left(x+y+z\right)}^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]\\ & =& \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}+2\left[\sqrt{\frac{1}{c^2}}+\sqrt{\frac{1}{a^2}}+\sqrt{\frac{1}{b^2}}\right]\\ & =& \left(\frac{a^2+ b^2+c^2}{abc}\right)+2\left[\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right]\\ & =& \frac{1}{abc}\left(a^2+ b^2+c^2\right)+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
 
Hence, the correct answer is option (b).