$\frac{d}{dx}\:y=\left(3x^3-1\right)\left(2x+5\right)^3$
$f\left(x\right)=\left(2x^3+7\right)^4\left(4x^3-5\right)^5$
$\left(3w-u\right)^3$
$\int\:\frac{2x^3-x^2+2}{x^2-x}dx$
$\frac{x^3-2x^2-4x+8}{x+2}$
$-16-\left(-11\right)$
$2x^3+3y^2=5$
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