$\frac{dy}{dx}=-y=12e^{6x}$
$\left(4x^2+3y^3+\frac{1}{9}z\right)^2$
$\left(\frac{2x^2+3}{4x+2}\right)$
$\frac{d}{dx}x^2-3y^2+7=0$
$\left(2x+\frac{1}{2}\right)\left(3x+\frac{1}{3}\right)$
$\frac{x^3+3}{x+3}$
$16z^8-121x^2y^6z^{12}$
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