$\left(4x^3\right)^2\:$
$\left(x-4\right)\left(x+1\right)$
$4x^2+20x-9$
$-3\csc\left(x\right)=4$
$\frac{12}{x}+x-8=0$
$y'+2y=2$
$67\cdot5\cdot y$
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