5. In $\triangle A B C$, $A B=6, B C=4$, and the length of the median on side $A C$ is $\sqrt{10}$. Then the value of $\sin ^{6} \frac{A}{2}+\cos ^{6} \frac{A}{2}$ is
5. $\frac{211}{256}$.

Let $M$ be the midpoint of $AC$. By the median length formula, we have
$$
\begin{array}{l}
4 B M^{2}+A C^{2}=2\left(A B^{2}+B C^{2}\right) \\
\Rightarrow A C=\sqrt{2\left(6^{2}+4^{2}\right)-4 \times 10}=8 .
\end{array}
$$

By the cosine rule, we get
$$
\begin{array}{l}
\cos A=\frac{C A^{2}+A B^{2}-B C^{2}}{2 C A \cdot A B}=\frac{7}{8} . \\
\text { Hence } \sin ^{6} \frac{A}{2}+\cos ^{6} \frac{A}{2} \\
=\left(\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}\right) \cdot \\
\quad\left(\sin \frac{A}{2}-\sin ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{A}{2}+\cos ^{4} \frac{A}{2}\right) \\
=\left(\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}\right)^{2}-3 \sin ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{A}{2} \\
=1-\frac{3}{4} \sin ^{2} A=\frac{1}{4}+\frac{3}{4} \cos ^{2} A \\
=\frac{211}{256} .
\end{array}
$$