$\frac { x ^ { 2 } + 1 } { x + 2 } = 0$
$8\left(x-1\right)^3$
$3x-2<8$
$\left(4x-\frac{5}{2}\right)\:\left(4x+\frac{5}{2}\right)$
$\left(2+5\right)\left(2+5\right)^2$
$9ax-\:6by-3bx+3ay$
$\left(4abc-1\right)^2$
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