8. $(8,9)$ It is known that the number $\frac{(2+\sqrt{3})^{4}+(2-\sqrt{3})^{4}}{(2+\sqrt{3})^{6}+(2-\sqrt{3})^{6}}$ is rational. Write this number as an irreducible fraction.
Answer: $\frac{97}{1351}$.

Solution: Let $a=2+\sqrt{3}, b=2-\sqrt{3}$. Notice that $a+b=4$, $a b=1$, therefore, $a^{2}+b^{2}=(a+b)^{2}-2 a b=14$. Similarly, we find $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2(a b)^{2}=194$ and $a^{6}+b^{6}=\left(a^{2}+\right.$ $\left.b^{2}\right) \cdot\left(a^{4}-a^{2} b^{2}+b^{4}\right)=2702$. By simplifying the fraction $\frac{194}{2702}$, we get $\frac{97}{1351}$.