$3b^3\cdot2b^3$
$\left(7x^3+y+4\right)^2$
$-3\:+\:\left[3\:-\:\left(4\:-\:9\right)\right]\:-\:6\:-\left[8\:\cdot\:\left(3\:\cdot\:2\:+\:1\right)\right]$
$\frac{\left(x-1\right)}{\left(3x^3-4x^2-x+2\right)}$
$2500\left(1+\frac{x}{1}\right)^2$
$y^6+7y^3-8$
$-3x-2>3x+8$
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