$\frac{6^9}{6^3}$
$-7x-8\ge-9+9x$
$\left(-12\right)\cdot\left(-9\right)\cdot\left(14\right)$
$\left(2x-3\right)\left(x+7\right)$
$60n^3q^2\:-\:36n^5q^2\:-\:72n^3q^3$
$\left(2x-q\right)^2$
$3x^5+4x^5$
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