$\frac{x^4+4x^3-4x^2-5}{-2}$
$x^{-3}y^2$
$-42\cdot7$
$\frac{2x^3-3x^2}{x}$
$\frac{dy}{dx}\left(3x^3+y=3y^3+x\right)$
$\left(\sqrt[3]{6x^3y^2}\right)\left(\sqrt[3]{2x^2y^5}\right)$
$4x^2=81$
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