$\left(8x^2y-4z\right)^3$
$100-x^2y^2$
$4\left(7\right)^2-4\left(7\right)-24$
$x^3=-\frac{x}{2}$
$\int-\frac{\left(\sin\left(3x\right)\right)}{\left(\cos\left(3x\right)^3\right)}dx$
$\tan\left(x\right)=\arcsin\left(16x\right)$
$sin2\left(x\right)\:+\:4\:cos\left(x\right)\:+\:4$
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