$\left(\frac{1}{\sin^2x}-\frac{1}{\tan^2x}\right)=\frac{1-\cos}{1+\cos}$
$\frac{16y^7-24y^5}{8y^2}$
$3+\left(-2\right)-\left(-2\right)+\left(-4\right)$
$6\left(7\:-\:x\right)=\:8\left(6-x\right)$
$-680+\left(-47\right)$
$\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^{2\:}+x\:+1\right)$
$6x^2-3x+4$
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