$\frac{x^4-4x^3+3x^2+4x-4}{x-1}$
$.8\:-\:\left(-3\right)\:$
$y'=\frac{4y}{x\left(y-3\right)}$
$\frac{cosxcotx}{cotx-cosx}$
$-7a+11b-8a-1a$
$\frac{1}{1+4x+4x^2}$
$\left(2x^4y^{-\frac{4}{5}}\right)^3\left(8y^2\right)^{\frac{2}{3}}$
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