$9x-6y\:=\:2$
$\left(-3x^2+3x\right)^3$
$\frac{dy}{dx}=y\tan\left(x\right)+\sin\left(2x\right)$
$x^2+32x-8=0$
$\left(5x-9\right)\left(7x-17\right)$
$\left(2x+5\right)\left(x-4\right)=0$
$\frac{-4a^3}{3}+\frac{5b^2}{7}$
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