$\int x\left(1+x^2\right)\frac{3}{2}dx$
$\frac{1}{x}+\frac{1}{y}=\frac{4}{xy}$
$\left(10a^5+96b^4-88c^5\right)^2$
$\left(b-3\right)\left(b+8\right)$
$in\:3-\:in\left(x+5\right)\:-\:inx\:=\:0$
$a^2-80a+16$
$\frac{dy}{dx}=\cos\left(x\right)e^{y+\sin\left(x\right)}$
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