Step 1: For a Bernoulli equation $\frac{dy}{dx}+P(x)y=Q(x)y^n$, set $y=v^{1/(1-n)}$ directly, using the general exponent rule rather than dividing first.
Step 2: With $n=3$, this gives $y=v^{-1/2}$, that is $v=y^{-2}$.
Step 3: Differentiate: $\frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}$.
Step 4: Substitute $y=v^{-1/2}$ and this derivative into the original equation: \[-\frac{1}{2}v^{-3/2}\frac{dv}{dx}+P(x)v^{-1/2}=Q(x)v^{-3/2}.\]
Step 5: Multiply through by $-2v^{3/2}$: \[\frac{dv}{dx}-2P(x)v=-2Q(x),\] a linear first-order equation in $v$, confirming the same substitution by a direct route.
\[\boxed{v=y^{-2}}\]