$\sqrt[3]{ab^2}$
$\lim_{x\to4}\left(\frac{x^2-4x}{e^{2x-8}-1}\right)$
$x+2\le0$
$\frac{3}{x\left(x+3\right)}$
$17\:-\:6\:+\:4\:-\:12\:+\:3\:-\:4\:-\:9\:+\:8$
$7\left(2\right)-9$
$\left(3x-6\right)^3$
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