$\frac{\left(3-x\right)\left(x^2+6x-7\right)}{\left(x-2\right)^2-1}$
$-5\cdot3\left(\left(-4-2\right)\right)$
$\frac{a^2}{2b}+\frac{2b^2}{a}$
$\left(x^2y^3-5\right)\left(x^2y^3-12\right)$
$\left(4^2\right)\left(4^{-4}\right)$
$3.851+38$
$\lim_{b\to\infty}b^3$
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