$\left(2y^3-9\right)^2$
$y^3-12y^2+36y=0$
$\sec y-\tan y=\frac{\cos y}{1+\sin y}$
$\left(4^6\cdot4^4\cdot4^2\right)$
$5\cdot4-2\cdot8-3$
$\lim_{x\to8}2$
$\left(3+3\sin\left(y\right)\right)\left(3+3\sin\left(-y\right)\right)=9\cos^2\left(y\right)$
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