$\left(2x^3+4y^2\:\right)^3$
$-2\left(0\right)^2+4x-2$
$x^2+30x-225$
$\int_{-4}^0\left(\frac{1}{\sqrt{16-x^2}}\right)dx$
$\left(+9\right)x\:-36$
$\frac{d}{dx}\left(x^x\right)=\ln\left(x\right)^{\sin\left(x\right)}$
$\left(2x-3\right)\left(5x+8\right)=12$
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