$\frac{m^3}{4m^8}$
$\sec x+\cos^2x$
$3y-4y^2+3y$
$a\left(4a^2-4ab^2+b^2\right)-b\left(4a^2-4ab+b^2\right)$
$x^2+x-2=x-1$
$xy\frac{dy}{dx}-y-2=0$
$2+\left(4-2\right)\cdot3$
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