$\left(x+2\right)\left(x+3\right)=8$
$x^8+x^6+x^4$
$\left(3a^2+4ab+4b^2\right)\left(2a+3b\right)$
$y^2+2x=2$
$b^2+7b=8$
$-\left(\frac{-11}{1}\right)$
$\lim_{x\to\infty}\left(\frac{x^7-6}{x^6+12}\right)$
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