$\left(3x^2-7y^3\right)\:^3$
$5^3.5$
$-9\cdot\left(-12+4\right)$
$\frac{1}{\cos\left(x\right)}-\frac{\cos\left(x\right)}{1+\sin\left(x\right)}=\tan\left(x\right)y$
$x^3-9x^2$
$\frac{x}{y\left(x^2-1\right)}=\frac{dy}{dx}$
$\frac{\left(-1\right)^nx^n}{\left(2^n\right)\cdot n}$
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