$\left(x+4\right)^2\left(x-4\right)^2-2\:\left(x^2-4\right)$
$0.36\frac{1}{2}$
$cos=-\frac{2}{5}$
$v^2+8vw+16w^2$
$2ax+5bx+2ay+5by$
$3\left(8\right)^4-3\left(8\right)^3-\left(8\right)^2+3\left(8\right)-1\:$
$\lim_{x\to\infty}\left(1+4x\right)^{\frac{1}{2x}}$
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