RD Sharma 2020 solution class 9 chapter 5 Factorization of Algebraic Expressions Exercise 5.4

Exercise 5.4

Page-5.23

Question 1:

Factorize each of the following expressions:

a³ + 8b³ + 64c³ − 24abc
 

Answer 1:

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.

Question 2:

x³ − 8y³ + 27z³ + 18xyz

Answer 2:

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.

Question 3:

27x³ − y³ − z³ − 9xyz

Answer 3:

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 .

Question 4:

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

Answer 4:

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-5.24

Question 5:

8x³ +27y³ − 216z³ + 108xyz

Answer 5:

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 .

Question 6:

125 + 8x³ − 27y³ + 90xy

Answer 6:

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 .

Question 7:

8x³ − 125y³ + 180xy + 216

Answer 7:

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 .

Question 8:

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

(ii) x² + 4y² + z³ + 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)

Answer 8:

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

$$\begin{gathered} \left(x^2+4y^2+z^2+2xy+xz-2yz\right)\left(-z+x-2y\right) \\ =\left(-z\right)\left(x^2+4y^2+z^2+2xy+xz-2yz\right)+\left(x\right)\left(x^2+4y^2+z^2+2xy+xz-2yz\right)+\left(-2y\right)\left(x^2+4y^2+z^2+2xy+xz-2yz\right) \\ =-x^{2}z-4y^{2}z-z^3-2xyz-x{z}^2+2y{z}^2+x^3+4x{y}^2+x{z}^2+2x^{2}y+x^{2}z-2xyz-2x^{2}y-8y^3-2y{z}^2-4x{y}^2-2xyz+4y^{2}z \\ =x^3-8y^3-z^3-x^{2}z+2x^{2}y+x^{2}z-2x^{2}y-4y^{2}z+4x{y}^2-4x{y}^2+4y^{2}z-x{z}^2+2y{z}^2+x{z}^2-2y{z}^2-2xyz-2xyz-2xyz \\ =x^3-8y^3-z^3-6xyz \end{gathered}$$

Hence, the required value is $x^3-8y^3-z^3-6xyz.$

Question 9:

(3x − 2y)³ + (2y − 4z)³ + (4z − 3x)³

Answer 9:

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.

Question 10:

(2x − 3y)³ + (4z − 2x)³ + (3y − 4z)³

Answer 10:

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 .

Question 11:

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

Answer 11:

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

Question 12:

(a − 3b)³ + (3b − c)³ + (c − a)³

Answer 12:

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.

Question 13:

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

Answer 13:

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 .

Question 14:

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

Answer 14:

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.

Question 15:

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

Answer 15:

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 .

Question 16:

(x – 2y)³ + (2y – 3z)³ + (3z – x)³

Answer 16:

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


 

Question 17:

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

Answer 17:

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

Question 18:

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

Answer 18:

$$\begin{gathered} \mathrm{Given}: a+b+c=0 \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that}, \\ \mathrm{if} a+b+c=0, \mathrm{then} a^3+b^3+c^3=3abc \\ \mathrm{Therefore}, a^3+b^3+c^3=3abc \\ \mathrm{Dividing} \mathrm{by} \left(abc\right) \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{ a^3+b^3+c^3}{abc}=\frac{3abc}{abc} \\ \Rightarrow \frac{ a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=3 \\ \Rightarrow \frac{ a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3 \\ \mathrm{Hence}, \frac{ a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3. \end{gathered}$$