$4^2-4\left(3\right)\left(4\right)$
$\left(\frac{3x}{\frac{1}{y^3}}\right)^3$
$\frac{dy}{dx}=\frac{\left(x^5+y^5\right)}{xy^4}$
$-9+3\left(11+21\right)-32$
$2^2-1$
$\left(x+1\right)^2=\left(x+1\right)^2+4x$
$x^2\:dy+y^2dx$
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