$\lim_{x\to4}\left(\frac{x^2-64}{x^2+4x-36}\right)$
$\left(3a^3-4b^5\:\right)^3$
$x^3+y^3=8$
$x\left(x+7\right)\left(x+3\right)$
$x^2-10x-9=0$
$\frac{7}{9\:}\:x\:-\:\frac{5}{3}\:>-2$
$\frac{2}{-2}$
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