Question

$\frac{3}{4}(a+y)[y+a-\frac{1}{3}(y+a-\frac{1}{4}(a+y))]$ is equal to

• A. ${(a+y)}^2$
• B. $\frac{3a}{16}$
• C. $\frac{9}{16}{(a+y)}^2$
• D. $1$

Answer 1

The correct option is C $\frac{9}{16}{(a+y)}^2$

$\frac{3}{4}(a+y)[y+a-\frac{1}{3}(y+a-\frac{1}{4}(a+y\left)\right)]$

=$\frac{3}{4}(a+y)\left[\right(a+y)-\frac{1}{3}(a+y)+\frac{1}{12}(a+y\left)\right]$

=$\frac{3}{4}(a+y)(a+y)(1-\frac{1}{3}+\frac{1}{12})$

=$\frac{3}{4}\times \frac{9}{12}(a+y{)}^2$=$\frac{9}{16}(a+y{)}^2$