Pre calculus fraction simplify question

Question

Simplify:

$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$

Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer

Answer 1

Let $u=\frac{2x}{3}$ and notice that $u^{4}=\big(\frac{2x}{3}\big)^{4}=\frac{2^{4}x^{4}}{3^{4}}=\frac{16x^{4}}{81}$.

So we get:

$\frac{\frac{16x^{4}}{81}-y^{4}}{\frac{2x}{3}+y}=\frac{u^{4}-y^{4}}{u+y}=\frac{(u^{2}-y^{2})(u^{2}+y^{2})}{u+y}=\frac{(u-y)(u+y)(u^{2}+y^{2})}{u+y}=(u-y)(u^{2}+y^{2})=(\frac{2x}{3}-y)(\frac{4x^{2}}{9}+y^{2})$

$=(\frac{1}{3}(2x-3y))(\frac{1}{9}(4x^{2}+9y^{2}))=\frac{1}{27}(2x-3y)(4x^{2}+9y^{2})=\frac{1}{27}(8x^{3}-12x^{2}y+18xy^{2}-27y^{3})$

Answer 2

hint: try multiplying by $\frac{\frac{2x}{3}-y}{\frac{2x}{3}-y}$ and then $\frac{\frac{4x^2}{9}+y^2}{\frac{4x^2}{9}+y^2}$

Answer 3

Multiplying by 81 on the top and the bottom gives

$\displaystyle\frac{16x^4-81y^4}{27(2x+3y)}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{27(2x+3y)}=\frac{(2x-3y)(2x+3y)(4x^2+9y^2)}{27(2x+3y)}=$

$\frac{1}{27}(2x-3y)(4x^2+9y^2)$.

Answer 4

$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}=\frac{16x^{4}-3^4y^4}{81}\cdot \frac{3}{2x+3y}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{81}\cdot \frac{3}{2x+3y}=\frac{(2x-3y)(2x+3y)(4x^2+3y^2)}{81}\cdot \frac{3}{2x+3y}.$$