$\left(x+8\right)\left(x-8\right)^2$
$2x^4+14x^3+8x^2$
$m^6-8n^{12}$
$y'\:+\:2x^2\:+\:3y\:=\:0$
$\lim_{x\to\infty}\left(\frac{2x^2}{e^{8x}}\right)$
$60+-1$
$4\ln\left(x+2\right)+6\ln\left(x+3\right)-\ln\left(x+1\right)-6\ln\left(x+2\right)$
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