$2x\left(x+2\right)^2-8\left(x^2+x\right)$
$\left(2x^2\right)^3x+3\left(x-3\right)$
$2x+11=6x-1$
$-6xy-8+5x^2+6xy-9$
$5z-8+2z+12-12z$
$64f-64g+96$
$\lim_{x\to-3}\left(\frac{\left(x^2+12x+17\right)}{-18-3x+x^2}\right)$
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