Question

Find the square of:

$\frac{3a}{2b}-\frac{2b}{3a}$.

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

Answer 1

The correct option is D $\frac{9a^2}{4b^2}-2+\frac{4b^2}{9a^2}$

Given, $\frac{3a}{2b}-\frac{2b}{3a}$.

On squaring, we get,

$={(\frac{3a}{2b}-\frac{2b}{3a})}^2$.

We know, $(a-b{)}^2=a^2-2ab+b^2$.

Then,

$={\left(\frac{3a}{2b}\right)}^2-2\left(\frac{3a}{2b}\right)\left(\frac{2b}{3a}\right)+{\left(\frac{2b}{3a}\right)}^2$

$=\frac{9a^2}{4b^2}-2+\frac{4b^2}{9a^2}$.

Therefore, option $D$ is correct.