Question

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 1

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 .