$\left(2s^2-3t^3\right)^3$
$3\tan x+4=7$
$5x+6\left(3x^2-4x+8\right)$
$\frac{dx}{dv}=\frac{2}{x}$
$-\left(-5\right)+\left(-4\right)+\left(+6\right)-3$
$\left(3b+5c\right)\left(3b-5c\right)$
$\sqrt{1}^{-1}$
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