$\left(3x+4y\right)\left(3x-4y\right)=9x^2-16y^2$
$\frac{1}{2}\cdot\left(128-84\right)+128$
$\sqrt{2}\cdot\sqrt[2]{2}$
$-8\:\frac{1}{4}$
$6a-3a-a-12$
$\int\frac{3x^3-4x^2+5x+6}{3x+2}dx$
$-\frac{1}{6}x=-5$
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