$\left(3y-4z\right)^3$
$48=-3x$
$x^2+10x+-25$
$\frac{x^2-4}{\left(3x^3-24\right)}$
$x\left(x-4\right)=0$
$xydx+\left(x^2+y^2\right)dy=0$
$a^2+b^2-2ab\:.\:a-b$
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