$\frac{cos\theta\cdot sec\theta\:}{1+tan^2\theta\:}=cos^2\theta\:$
$1^{\frac{2}{3}}$
$\frac{w^8}{w^7}\cdot\frac{w^5}{w^2}$
$b^m\cdot b^n$
$x^2-6-9x$
$8ax+2a-4bx-b$
$4\left(4\right)\left(-4\right)\left(-4\right)$
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