$r^2-6r+9$
$\frac{\left(1+tan^2\left(b\right)\right)}{tan^2\left(b\right)}$
$-9\:+4\:\cdot\:-27$
$9\left(-8+6\right)+9-4\left(7-3\right)$
$2\left(3\right)^2\left(8\right)^2$
$\frac{9x^3+7x^2-3x}{x-10}$
$\left(5x^4y^3+1\right)^2$
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