Problem 1198

4. Given that $a, b, c$ are the roots of the equation $x^{3}+(a b+b c+c a) x-a b c=0$.

Let $T_{n}=a^{n}+b^{n}+c^{n}(n=0,1,2, \cdots)$, then $T_{a}=-(a b+b c+c a) T_{a-2}+a b c T_{a-3}(n \geqslant 3)$.
Also, $T_{0}=3, T_{1}=0$, so $T_{3}=-(a b+b c+c a) T_{1}+b a c T_{0}=3 a b c$.
And $T_{2}=a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=-2(a b+b c+c a)$, thus
$$\begin{array}{l} T_{n}=\frac{1}{2} T_{2} T_{n-2}+\frac{1}{3} T_{3} T_{n-3}(n \geqslant 3), \text { so } T_{4}=\frac{1}{2} T_{2}^{2}, \\ T_{5}=\frac{1}{2} T_{2} T_{3}+\frac{1}{3} T_{3} T_{2}=\frac{5}{6} T_{2} T_{3}, T_{7}=\frac{1}{2} T_{2} T_{5}+\frac{1}{3} T_{3} T_{4}=\frac{7}{12} T_{2}^{2} T_{3}, \end{array}$$

Thus $T_{2} T_{3} T_{4} T_{5}=\frac{5}{12} T_{2}^{4} T_{3}^{2}, T_{7}^{2}=\frac{49}{144} T_{2}^{4} T_{3}^{2}$.
Therefore, $\frac{T_{7}^{2}}{T_{2} T_{3} T_{4} T_{5}}=\frac{49}{144} \cdot T_{2}^{4} T_{3}^{2} \cdot \frac{12}{5 T_{2}^{4} T_{3}^{2}}=\frac{49}{60}$, which proves the required equation.