RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities Exercise 4.4

Exercise 4.4

Page-4.24

Question 1:

Find the following products:

(i) (3x + 2y) (9x² − 6xy + 4y²)

(ii) (4x − 5y) (16x² + 20xy + 25y²)

(iii) (7p⁴ + q) (49p⁸ − 7p⁴q + q²)

(iv) $\left(\frac{x}{2}+2y\right) \left(\frac{x^2}{4}-xy + 4y^2\right)$

(v) $\left(\frac{3}{x}-\frac{5}{y}\right) \left(\frac{9}{x^2}+\frac{25}{y^2}+\frac{15}{xy}\right)$

(vi) $\left(3+\frac{5}{x}\right) \left(9-\frac{15}{x}+\frac{25}{x^2}\right)$

(vii) $\left(\frac{2}{x}+3x\right) \left(\frac{4}{x^2}+9x^2-6\right)$

(viii) $\left(\frac{3}{x}-2x^2\right) \left(\frac{9}{x^2}+4x^4-6x\right)$

(ix) (1 − x) (1+ x + x²)

(x) (1 + x) (1 − x + x²)

(xi) (x² − 1) (x⁴ + x² + 1)

(xii) (x³ + 1) (x⁶ − x^{3 +} 1)

Answer 1:

(i) In the given problem, we have to find the value of

 Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(ii) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(iii) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(iv) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(v) Given

We shall use the identity

We can rearrange the as

$$\begin{gathered} =\left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)-\left(\frac{5}{y}\right)\times \left(\frac{5}{y}\right)\times \left(\frac{5}{y}\right) \\ =\frac{27}{x^3}-\frac{125}{y^3} \end{gathered}$$

Hence the Product value of is

(vi) Given

We shall use the identity ,

we can rearrange the as

Hence the Product value of is

(vii) Given

We shall use the identity,

We can rearrange the as

Hence the Product value of is

(viii) Given

We shall use the identity

We can rearrange the as

$$\begin{gathered} \left(\frac{3}{x}-2x^2\right)\left({\left(\frac{3}{x}\right)}^2+{\left(2x^2\right)}^2-\left(\frac{3}{x}\right)\left(2x^2\right)\right) \\ ={\left(\frac{3}{x}\right)}^3-{\left(2x^2\right)}^3 \\ =\left(\frac{3}{x}\right)\left(\frac{3}{x}\right)\left(\frac{3}{x}\right)-\left(2x^2\right)\left(2x^2\right)\left(2x^2\right) \\ =\frac{27}{x^3}-8x^6 \end{gathered}$$

Hence the Product value of is

(ix) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(x) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(xi) Given

We shall use the identity

We can rearrange the as

$\left(x^2-1\right)\left[{\left(x^2\right)}^2+\left(x^2\right)\left(1\right)+{\left(1\right)}^2\right]$

Hence the Product value of is

(xii) Given

We shall use the identity,

We can rearrange the as

Hence the Product value of is .

Question 2:

If x = 3 and y = − 1, find the values of each of the following using in identify:

(i) (9y² − 4x²) (81y⁴ +36x²y² + 16x⁴)

(ii) $\left(\frac{3}{x}-\frac{x}{3}\right) \left(\frac{x^2}{9}+\frac{9}{x^2}+1\right)$

(iii) $\left(\frac{x}{7}+\frac{y}{3}\right) \left(\frac{x^2}{49}+\frac{y^2}{9}-\frac{xy}{21}\right)$

(iv) $\left(\frac{x}{y}-\frac{y}{3}\right) \frac{x^2}{16}+\frac{xy}{12}+ \frac{y^2}{9}$

(v) $\left(\frac{5}{x}+5x\right)$ $\left(\frac{25}{x^2}-25+25x^2\right)$

Answer 2:

In the given problem, we have to find the value of equation using identity

(i) Given 

We shall use the identity 

We can rearrange the as

Now substituting the value  in we get,

Hence the Product value of is 

(ii) Given

We shall use the identity 

We can rearrange the as

$$\begin{gathered} =\left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)-\left(\frac{x}{3}\right)\times \left(\frac{x}{3}\right)\times \left(\frac{x}{3}\right) \\ =\frac{27}{x^3}-\frac{x^3}{27} \end{gathered}$$

Now substituting the value  in we get,

Hence the Product value of is 

(iii) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in 

Taking Least common multiple, we get 

Hence the Product value of is 

(iv) Given

We shall use the identity 

We can rearrange the as

$$\begin{gathered} =\left(\frac{x}{4}\right)\times \left(\frac{x}{4}\right)\times \left(\frac{x}{4}\right)-\left(\frac{y}{3}\right)\times \left(\frac{y}{3}\right)\times \left(\frac{y}{3}\right) \\ =\frac{x^3}{64}-\frac{y^3}{27} \end{gathered}$$

Now substituting the value  in we get,

Taking Least common multiple, we get 

Hence the Product value of is 

(v) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in 

Taking Least common multiple, we get 

​

Hence the Product value of is .

Page-4.25

Question 3:

If a + b = 10 and ab = 16, find the value of a² − ab + b² and a² + ab + b²

Answer 3:

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

Given

We shall use the identity ${\left(a+b\right)}^3=a^3+b^3+3ab(a+b)$

We can rearrange the identity as 

Now substituting values in as,

We can write as  

Now rearrange as

Thus

Now substituting values

Hence the value of is respectively.

Question 4:

If a + b = 8 and ab = 6, find the value of a³ + b³

Answer 4:

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

Given

We shall use the identity

Hence the value of is .

Question 5:

If a + b = 6 and ab = 20, find the value of a³ − b³

Answer 5:

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

Given

We shall use the identity 

Hence the value of is .

Question 6:

 If x = −2 and y = 1, by using an identity find the value of the following

(i) 4y^{2 −} 9x² (16y⁴ + 36x²y²+81x⁴)

(ii) $\left(\frac{2}{x}-\frac{x}{2}\right) \left(\frac{4}{x^2}+\frac{x^2}{4}+1\right)$

(iii) $\left(5y+\frac{15}{y}\right) \left(25y^2-75+\frac{225}{y^2}\right)$

Answer 6:

(i) In the given problem, we have to find the value of using identity

Given

We shall use the identity

We can rearrange the as

                                                        $$\begin{gathered} =\left(4y^2\right)\times \left(4y^2\right)\times \left(4y^2\right)-\left(9x^2\right)\times \left(9x^2\right)\times \left(9x^2\right) \\ =64y^6-729x^6 \end{gathered}$$
Now substituting the value in we get,

Taking 64 as common factor in above equation we get,

Hence the Product value of is

(ii) In the given problem, we have to find the value of using identity

Given

We shall use the identity

We can rearrange the as

                                   $$\begin{gathered} =\left(\frac{2}{x}\right)\times \left(\frac{2}{x}\right)\times \left(\frac{2}{x}\right)-\left(\frac{x}{2}\right)\times \left(\frac{x}{2}\right)\times \left(\frac{x}{2}\right) \\ =\frac{8}{x^3}-\frac{x^3}{8} \end{gathered}$$

Now substituting the value in we get,

Hence the Product value of is = 0.

(iii) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in

Hence the Product value of is