Question

Find the product of $(\frac{1}{2}{x}^2-\frac{1}{3}{y}^2)$ and $(\frac{1}{2}{x}^2+\frac{1}{3}{y}^2)$.

• A. $\frac{1}{9}{y}^4-\frac{1}{4}{x}^4$
• B. $\frac{1}{4}{x}^4-\frac{1}{9}{y}^4$
• C. $4x^2-9y^4$
• D. $\frac{1}{4}{x}^4+\frac{1}{9}{y}^4$

Answer 1

The correct option is B $\frac{1}{4}{x}^4-\frac{1}{9}{y}^4$

Given, product of $(\frac{1}{2}{x}^2-\frac{1}{3}{y}^2)$ and $(\frac{1}{2}{x}^2+\frac{1}{3}{y}^2)$

i.e. $(\frac{1}{2}{x}^2-\frac{1}{3}{y}^2)\times (\frac{1}{2}{x}^2+\frac{1}{3}{y}^2)$.

We know, the identity $(a+b)(a-b)=a^2-b^2$.

Then,

$(\frac{1}{2}{x}^2-\frac{1}{3}{y}^2)\times (\frac{1}{2}{x}^2+\frac{1}{3}{y}^2)$

$=(\frac{1}{2}{x}^2{)}^2-(\frac{1}{3}{y}^2{)}^2$

$=\frac{1}{4}{x}^4-\frac{1}{9}{y}^4$.

Therefore, option $B$ is correct.