$\left(4a^2-3b^3\right)^2$
$\left(x+7\right)\left(x-3\right)+\left(x-2\right)^2$
$\frac{2x^3-7x^2-36x}{x+3}$
$3\left(2^2\right)-2\left(3\right)+4$
$\frac{3}{2}+x\ge2$
$\frac{2}{x}+\frac{1}{4}$
$17\sin\left(x\right)=0$
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