$x-8=10$
$\frac{2x^5y^{4\:}}{xy^2}$
$^{-3^2}$
$\left(-2x\right)^4\left(3x^2\right)^3$
$3x^2-6x-12=0$
$\int_1^{\infty}\left(\frac{1}{36+x^2}\right)dx$
$3x^3+6x+3$
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