$=\frac{\left(a+x\right)^2-y^2}{\left(a+x\right)-y}$
$\left(4x-z^3\right)\left(4x+z^3\right)$
$\:\frac{\left(6x^2+2\right)}{5x+6}$
$x^2-4x+32$
$x\:.\:\left(8\right)=-72$
$2sin^2\:\frac{x}{2}-1$
$\sqrt[3]{125\sqrt{32\sqrt[3]{8}}}$
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