<s>[INST] Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
 [/INST] 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:
$\frac{\frac{16x^{4}}{81}-y^{4}}{\frac{2x}{3}+y}=\frac{u^{4}-y^{4}}{u+y}=\frac{(u^{2}-y^{2})(u^{2}+y^{2})}{u+y}=\frac{(u-y)(u+y)(u^{2}+y^{2})}{u+y}=(u-y)(u^{2}+y^{2})=(\frac{2x}{3}-y)(\frac{4x^{2}}{9}+y^{2})$
$=(\frac{1}{3}(2x-3y))(\frac{1}{9}(4x^{2}+9y^{2}))=\frac{1}{27}(2x-3y)(4x^{2}+9y^{2})=\frac{1}{27}(8x^{3}-12x^{2}y+18xy^{2}-27y^{3})$
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