5. In $\triangle A B C$, $A B=6, B C=4$, 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 $\qquad$
By the median formula, let $AC = 2x$, we have $36 + 16 = 2(10 + x^2) \Rightarrow x^2 = 16 \Rightarrow AC = 2x = 8$. Thus, $\cos A = \frac{36 + 16 - 10}{2 \times 6 \times 4} = \frac{7}{8} \Rightarrow \cos^2 \frac{A}{2} = \frac{15}{16}, \sin^2 \frac{A}{2} = \frac{1}{16}$. Therefore, $\sin^6 \frac{A}{2} + \cos^6 \frac{A}{2} = \frac{15^3 + 1}{16^3} = \frac{15^2 - 15 + 1}{16^2} = \frac{211}{256}$.