$\frac{6}{1\:}^3$
$+289-412$
$\:-\:2\:x\:\le\:6\:$
$x^2-x-2>0$
$\frac{3x^2+10x+10}{x+2}$
$3\left(\frac{\sqrt[4]{3}}{\sqrt[2]{3}}\right).\left(\frac{\sqrt[4]{3}}{1}\right)$
$\frac{x^4+2}{x^2+5x+2}$
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