Question

Find the following product :
$(\frac{4}{3}x-\frac{3}{4}y)(\frac{4}{3}x-\frac{3}{4}y)$

• A. $\frac{16}{9}{x}^2+\frac{9}{16}{y}^2-2xy$
• B. $\frac{9}{16}{x}^2+\frac{16}{9}{y}^2-2xy$
• C. $9x^2+16y^2-2xy$

Answer 1

The correct option is A $\frac{16}{9}{x}^2+\frac{9}{16}{y}^2-2xy$

$(\frac{4}{3}x-\frac{3}{4}y)(\frac{4}{3}x-\frac{3}{4}y)=(\frac{4}{3}x-\frac{3}{4}y{)}^2$

On comparing with identity $(a-b{)}^2=a^2+b^2-2ab$ we get,

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

$\Rightarrow (\frac{4}{3}x-\frac{3}{4}y{)}^2=\frac{16}{9}{x}^2+\frac{9}{16}{y}^2-2xy$