$\left(-4\sqrt[3]{3a^2b^4}\right)$
$6-\left(-3\right)-1-\left(-6\right)$
$6\left(x+8\right)$
$\frac{125+x^6}{5+x^2}$
$6x^2-5x+6=0$
$\frac{cos^3x\:-\:3cos\:x\:sin^2x}{cos\:x}=cos^2x-3sin^2x$
$\lim_{x\to-\infty}\left(\frac{4x^2+3}{2x^2-1}\right)$
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