$x^2+4x-28$
$\left(-3\right)^4x\left(-3\right)^3$
$\lim_{x\to4}\left(\frac{\left(x^2-16\right)}{x^2-5x+6}\right)$
$\frac{dy}{dx}=\frac{\left(x^2+y^2\right)}{-2xy}$
$\left(m^2-2n^3\right)^3$
$2-\left[-64\right]$
$-4\left(9x^2-7\right)$
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