$\frac{dx}{dy}=\frac{1+2y^2}{y\cos x}$
$\left(x\right)^3+\left(3y\right)^3$
$-\frac{168}{\left(-7\right)}$
$\left(4b+5b\right)^3$
$\left(\sin\left(4b\right)+4\right)^2$
$-4\left(-3\right)\left(-3\right)\left(-4\right)\left(-4\right)$
$\pi\sqrt{-\pi}$
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