$\frac{1-tan^4x}{\sec^2x}=1-\tan^2x$
$x^2-18x+x$
$\left(x\:-\:3\right)\:\left(x^3\:+\:2x\:-\:4\right)$
$12y+5y+4y$
$\frac{r^2s^{-3}}{3s^2}\cdot\frac{s^2}{x^{-4}}$
$\frac{x-3x}{2+5x}$
$4y^2-8$
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