$\left(2x^3\right)\left(3x^5\right)\left(6x\right)^2$
$\left(4x+12\right)\left(4x+2\right)\left(5x+14\right)$
$0.25+1.75$
$\frac{2x+4}{5x}=\frac{2}{x}$
$81y^4-16x^2$
$\left(11u-3x^2y\right)\left(xy^4+11u\right)$
$\left(5y\:-\:3z\right)^3$
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