$\frac{1}{2}\left(6\right)^2\:-3$
$\left(-3u\:w^2\right)^4$
$6xe^{-2x}-\int\left(6e^{-2x}\right)du$
$\left(-3ab^2\:\right)^2$
$\left(1-x\right)^2\left(2-\frac{1}{2}x\right)^4$
$\frac{2x^3-4x^2-5x-3}{x-1}$
$3x+2\left(x-1\right)=6x-9$
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