$\left(11-t^2w^4\right)\left(11-t^2w^4\right)$
$\left(\cos\right)^23x$
$a^2b\:+\:3\:a^2b^2\:+\:a\:b^3$
$\frac{6\sqrt[3]{a^2x}}{ax}$
$4x\:2\:\cdot\:5x$
$y^2-4y=\cos3a$
$x\frac{dy}{dx}-y=x^3$
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