$3x^3+x^3+5x^3$
$9\left(m+5n\right)^2-4\left(m-n+mn\right)^2$
$\frac{\frac{3}{4}b^3}{\frac{2}{3}c^2}$
$4x+2>x+4$
$-15+7+6-4+12+8+3-2-1+9$
$x^2+9x+2$
$\left(\frac{5x}{3}+3\right)\cdot\left(\frac{5x}{3}-3\right)$
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