$0=3x^2$
$\tan\left(x\right)+\left(-1-\tan\left(x\right)^2\right)\tan\left(x\right)=-\tan\left(x\right)^3$
$-16t^2\:+128t\:+128$
$\left(x-3\right)\left(2x+1\right)\left(3x-2\right)$
$b3\:-\:125$
$\frac{a^3+g^3}{a+g}$
$\int\frac{e^{-x}}{\left(4e^{-2x}+1\right)^{\frac{3}{2}}}dx$
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