$\frac{2^{-3}}{4^{-2}}$
$\frac{x^{12}}{x^2}$
$-x^2+6=x^2-4x$
$\left(x+4\right)\left(x-2\right)+\left(x-6\right)\left(x+4\right)-2x^2$
$1+1-8$
$\left(x+1\right)^2-\left(x+y\right)^2$
$\lim_{x\to0}\frac{x^3-x^2-x-1}{x^2-1}$
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