3.366. $\sin ^{3} \alpha-\cos ^{3} \alpha$, if $\sin \alpha-\cos \alpha=n$.
Solution.

$\sin ^{3} \alpha-\cos ^{3} \alpha=(\sin \alpha-\cos \alpha)\left(\sin ^{2} \alpha+\sin \alpha \cos \alpha+\cos ^{2} \alpha\right)=$

$=[\sin \alpha-\cos \alpha=n]=n(1+\sin \alpha \cos \alpha)=\left[\sin \alpha \cos \alpha=\frac{1-n^{2}}{2}\right]=$

$=n\left(1+\frac{1-n^{2}}{2}\right)=\frac{n\left(2+1-n^{2}\right)}{2}=\frac{n\left(3-n^{2}\right)}{2}=\frac{3 n-n^{3}}{2}$.

Answer: $\frac{3 n-n^{3}}{2}$.