$3\sqrt{4ab^3}\sqrt[3]{16a^2}$
$\lim\:_{x\to\:3}\left(\frac{x+3}{\frac{1}{x}+\:\frac{1}{3}}\right)$
$\frac{d}{dx}x^2+y^2-4xy=0$
$\left(6x\right)^2-2\left(6x\right)\left(y^2\right)^2$
$x^2-16x+24=0$
$15x^2+7x=2$
$6am\left(2+a\right)$
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