$41^{\infty}$
$\left(5^2\:+7x+8\right)+\left(3x+9\right)$
$\left(3a^2+4ab+4b^2\right)\left(2a+3b\right)$
$3m-4n-4m+n$
$cos\theta\:=\frac{-3}{\sqrt{13}}$
$\int-8xe^{4x}dx$
$8m^2n^6-\left(-4m^2n^6\right)$
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