$9z+3z$
$6x+10-2x=42$
$\left(2x^3+6x\right)^2$
$\left(x^3+2\right)\left(x+3\right)$
$493+918$
$\left(14x\right)\left(2x^2\right)$
$\frac{21\left(x-4\right)}{5}+\frac{3x+1}{8}+32x\le\:\:-\frac{43\left(8x+5\right)}{3}-\frac{1}{6}$
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