$\frac{\left(3x^3-4\right)}{\left(x+1\right)}$
$\frac{4x}{3}=12$
$\:y^3-2y^2+2y-4$
$8-2x=4$
$\int_0^{\pi}\left(\pi\left(\sin\left(x\right)\right)^2\right)dx$
$6x+25-2x+3$
$m^2+12^2$
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