$\int\left(3x^3-3x^2-3x+3\right)dx$
$\left(-5x^3y^2\right)\cdot\left(\frac{1}{2}x^3\right)$
$\frac{dy}{dx}=4x^3-5x+3$
$\left(\sqrt[3]{6x^3y^2}\right)\left(\sqrt[3]{2x^2y^5}\right)$
$\frac{x^3+12x^2-20x+7}{x+1}$
$x^4-2^3-x^2+1$
$\int\frac{23}{xln4x}dx$
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