$-10\left(3n-4\right)$
$x-42=13$
$\frac{1}{6}x\ge-\frac{181}{3}$
$\left(x+4\right)\left(x-2\right)+\left(x-6\right)\left(x+4\right)-2x^3$
$6a+2b+24c$
$\frac{\left(6x^3-3x^2-5y^1-x^1\right)}{x^2+2x+4}$
$12x^5+42x^3+147x$
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