I think it was meant to be $ a=u^3-3Auv^2$, not $ u^3-3Avu^2$ $ s^3=a^2+Ab^2$ $ s^3=(a+b\sqrt{-A})(a-b\sqrt{-A})$ Note $ 2$ cannot divide $ s$, since then $ a$ and $ b$ must both be even (look $ \mod 8$), but they are relatively prime. So $ 2$ cannot divide either $ a+b\sqrt{-A}$ or $ a-b\sqrt{-A}$. $ \gcd(a+b\sqrt{-A},a-b\sqrt{-A})=\\ \gcd(2a,a-b\sqrt{-A})=\\ \gcd(a,a-b\sqrt{-A})=\\ \gcd(a,-b\sqrt{-A})=1$ since $ a$ and $ b$ are relatively prime. Thus, $ a+b\sqrt{-A}$ and $ a-b\sqrt{-A}$ are relatively prime, and since their product is a cube, each must be a cube. So let $ (a+b\sqrt{-A})=(u+v\sqrt{-A})^3$ $ (a+b\sqrt{-A})=(u^3-3Auv^2+(3u^2 v-Av^3)\sqrt{-A})$ So $ a=u^3-3Auv^2$ and $ b=3u^2 v-Av^3$ Also note that $ s=(u+v\sqrt{-A})(u-v\sqrt{-A})=u^2+Av^2$, as desired.