4. Let $\sin \alpha-\cos \alpha=\frac{1}{3}$. Then $\sin 3 \alpha+\cos 3 \alpha=$ $\qquad$
4. $-\frac{25}{27}$.

Notice,
$$
\begin{array}{l}
2 \sin \alpha \cdot \cos \alpha \\
=\sin ^{2} \alpha+\cos ^{2} \alpha-(\sin \alpha-\cos \alpha)^{2} \\
\quad=1-\frac{1}{9}=\frac{8}{9} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } \sin 3 \alpha+\cos 3 \alpha \\
=\left(3 \sin \alpha-4 \sin ^{3} \alpha\right)+\left(4 \cos ^{3} \alpha-3 \cos \alpha\right) \\
=3(\sin \alpha-\cos \alpha)-4\left(\sin ^{3} \alpha-\cos ^{3} \alpha\right) \\
=1-4\left(\sin ^{3} \alpha-\cos ^{3} \alpha\right) \\
=1-\frac{4}{3}\left(\sin ^{2} \alpha+\sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha\right) \\
=1-\frac{4}{3}\left(1+\frac{4}{9}\right)=-\frac{25}{27} .
\end{array}
$$