$( 3 p ^ { 2 } - 5 q ) ^ { 2 }$
$\left(2x^3+3y\right)^5$
$\frac { ( 1 - x ) ^ { 3 } } { x ^ { 3 } }$
$\left(-3\right)+\left(-4\right)\left(-10\right)-\left(-4\right)$
$\int_{-1}^3\left(\frac{2}{x^3}\right)dx$
$\:\frac{1}{x^3+4x}$
$\frac{dy}{dx}\left(4x^3y^2\right)$
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