$\frac{1}{4}\int e^{-3x}\left(3x^2+2x+1\right)dx$
$x^2+10x+12$
$\left(3a^4b^5\right)^3$
$6x^3+16x^2+10x$
$3x^2+20x+25$
$\sqrt{2}\cdot sin^3\left(x\right)=sin^2\left(x\right)$
$\frac{4r^4}{\left(4r\right)^3}$
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