$\left(18x^3y^2z^2\right)\left(9x^3y^1z^2\right)$
$3a+\left(a+b-4c\right)-\left(3a+5b-3c\right)-\left(b-c\right)$
$11x>7x-28$
$\lim_{x\to-\frac{1}{2}}\left(\frac{6\left(-\frac{1}{2}\right)-1}{-\frac{1}{2}}\right)$
$3+9x-4$
$\frac{729x^3-512y^3}{9x-8y}$
$-9\cdot32$
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