$\left(2+3b\right)^2$
$\left(3y-4\right)\left(-2y^2+4y-3\right)$
$1\cdot11+2$
$\frac{\left(rq+\left(\frac{wq}{2}\right)\right)}{q}$
$-4=\sin^2x-5$
$\frac{6a^{3}b^{3}-9a^{2}b^{2}+3ab^{4}}{3ab^{2}}$
$-8x-3=-75$
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