$\frac{3}{6:2}$
$\left(\frac{m^3}{m^{-7}}\right)^{\frac{1}{16}}$
$4a\:^3\:x\:-\:4a\:^2\:b\:+\:3bm\:-\:3amx$
$\left(x^2+5\right)\left(x^2-7\right)$
$\lim_{x\to0}\left(\frac{1-\cos\left(x^3\right)}{\sin\left(x\right)^2}\right)$
$6x\:-\:45\:=\:9$
$3u^2+8u-3=0$
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