$\frac{\left(2x+1\right)^{2}}{\left(3x-1\right)^{3}}$
$f\left(x\right)=\left(x+8\right)^5\left(3x-2\right)^3$
$\left(4t^2+\:3t\right)^3$
$\:-4k-9\left(-8k+6\right)$
$\frac{4x^4y^6}{20x^4y^4}$
$9-\left(-3r\right)$
$2\ln\left(x-7\right)=6$
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