$\left(42\left(\frac{1}{2}z^8-\frac{1}{7}\right)\right)^3$
$\sqrt{4-x^2}\cdot\sqrt{x^2-4}$
$-42m^6n^4+49n^8$
$\frac{e^n-e^n}{e^n+e^{-n}}$
$3e^{4x}+6$
$\lim_{x\to\infty}\left(x-\sqrt{x^2-6x}\right)$
$\lim_{x\to-infinity}\left(z^2e^z\right)$
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