$f\left(x\right)=\left(3-2x\right)^5$
$\left(2x+7^2\right)$
$13y^2-7y\:-6$
$\left(\sqrt{7n}\right)^5$
$\left(x-1\right)^0$
$3\left(x-6\right)^{2}-\left(x+4\right)^{2}+4\left(x+8\right)\cdot\left(x-8\right)$
$\frac{3x+17}{-5}=-4$
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