$x-3<3x+6$
$\int\frac{3}{\left(x^3-x\right)}dx$
$\left(-15\right)\cdot\left(-6\right)$
$\left(11a^2-2\right)^3$
$z^2-9$
$\frac{3}{4}mnr-\frac{1}{3};\:m=5\:;\:n=10\:;\:r=6$
$\left(5x+3\right)+3x^2-2$
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