$\left(\sqrt[3]{5}x^2-x^3\right)\left(\sqrt[3]{25}x^4+\sqrt[3]{5}x^5+x^6\right)$
$\lim_{x\to0}\left(\frac{8x^2-1}{x\cdot\ln\left(x\right)}\right)$
$u^2+4u+8$
$\:y^2\:+\:8y\:+\:16.$
$xy+3x+y=6$
$2x-5<-12$
$\frac{64a^3b^5}{4a^3b^2}$
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