$\sqrt[3]{x^3y^6}$
$\:\left(4x^3+5y^4\right)^2$
$6\cdot3+16$
$x^2-3x-10\le0$
$\left(2x^3-3x^2+10x-6\right)\left(-2x\right)$
$\int\left(x^2\left(x^3-6\right)^{18}\right)dx$
$\left(y^2-12\right)\left(y^2+8\right)$
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