$-5\cdot19\cdot\left(-8\right)$
$b^2-8b+15$
$x+8>3x-4$
$2\left(x+3\right)\:^3$
$\lim_{x\to0}\left(\sqrt{9x^2+3x+1}-3x\right)$
$\frac{\sec\left(w\right)}{\tan\left(w\right)+\cot\left(w\right)}=\sin\left(w\right)$
$\left(x^2-4x+13\right)\left(x^2-4x+13\right)$
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