$\left(2x^2-\frac{1}{2}\right)^2$
$b^3-1$
$\frac{2}{3}\sqrt[3]{4m^2}.\frac{3}{4}\sqrt[5]{16m^4n}$
$\left(2x-1\right)\left(3x+2\right)$
$\left(-8\right).\left(-3\right)$
$5a+a-6$
$-2\left(y+2\right)+y$
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