$\left(\frac{2}{3}+3y^3\right)\left(3y^3-\frac{4}{5}\right)$
$x^2-14x+33\:$
$\frac{-11}{16mn}-\frac{7}{16mn}$
$3-5\cdot\left(2+6\cdot4\right)$
$\left(3x^2-y\right)^4$
$342\cdot21-134$
$-1\:\left(23\right)+5$
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