$2\:tan^2\:a\:+\:tan\:a=\:0$
$\frac{6}{8}\cdot\frac{4}{3}$
$2a^2x+6x^2$
$\frac{+2x^3+x^2-12x\:+7\:}{x-3}$
$\left(-7\right)-\left(-14\right)$
$\left(a\:+\:\sqrt{3}\right)\left(\sqrt{3}\:-\:a\right)$
$\left(\frac{3^4}{3^2}\right)\left(\frac{3^2}{3^3}\right)\left(3^5\right)$
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