$x-4=16$
$\left(x^4-3x\right)\left(x^4+2\right)$
$-16-\left(-7\right)$
$\left(\frac{\:x}{2\:}+\frac{2}{3}\right)\left(\frac{2x}{3}-\frac{2}{5}\right)$
$\left(5+a\right)\left(5-b\right)$
$\frac{1}{x+6}-\frac{x^2-3x-18}{x^2+12x+36}=\frac{4}{x+6}$
$\left(x+16\right)\left(x+24\right)$
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