$\tan^2\left(x\right)\sec\left(x\right)+sec\left(x\right)=sec^3x$
$x^3-9\:x-27+27$
$\frac{8a^3-43a^2-25a-30}{a-6}$
$\lim_{x\to\infty}\left(\frac{9x-8}{6+2\sqrt{x}}\right)$
$4r+14-r-6$
$-9x-9y$
$4.\left(-1\right)$
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