$\left(2xy^2-3z^4\right)^3$
$\frac{\sin\left(2x\right)}{\cos\left(x\right)}=1$
$19^2$
$3xy^{3}-18xyz-y^{2}+6z$
$\:21\:+\:1-2$
$\left(-14.\left(-26\right)\right)$
$16x^6-2x^3y^2+\frac{y}{16}^2$
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