$14-1-3$
$\left(4z-8\right)\left(4z+8\right)$
$81x^2-36xy+4y^2$
$-3a^xb^nc\left(-69^3b^4c\right)$
$\lim_{x\to3}\left(\frac{\left|x^2-x-6\right|}{x-3}\right)$
$-40x\:-\:35$
$4x^2+23x+33=2x+8$
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