Question

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

• A. $\frac{4x}{3y}$
• B. $2(4x^2+\frac{1}{9y^2})$
• C. $\frac{8x}{3y}$
• D. $\frac{4y}{3x}$

Answer 1

The correct option is C $\frac{8x}{3y}$

We have,

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

We know,

$a^2-b^2=(a+b)(a-b)\therefore {(2x+\frac{1}{3y})}^2-{(2x-\frac{1}{3y})}^2=(2x+\frac{1}{3y}+2x-\frac{1}{3y})(2x+\frac{1}{3y}-(2x-\frac{1}{3y}))=\left(4x\right)(2x+\frac{1}{3y}-2x+\frac{1}{3y})=\left(4x\right)\left(\frac{1+1}{3y}\right)=\left(4x\right)\left(\frac{2}{3y}\right)=\frac{8x}{3y}$