8. A uniform cube die has the numbers $1,2, \cdots, 6$ marked on its faces. Each time two identical dice are rolled, the sum of the numbers on the top faces is the point value of that roll. What is the probability that the product of the 3 point values obtained from 3 rolls is divisible by 14? (Express your answer as a simplified fraction.)

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The product of the points can be divisible by 14 if and only if 7 and an even number appear simultaneously among the 3 points. The probability of the points being 7, a non-7 odd number, and an even number in each roll is $\frac{1}{6}$, $\frac{1}{3}$, and $\frac{1}{2}$, respectively. When rolling 3 times, there are 27 possible combinations of the 3 points, among which 7, non-7 odd numbers, and even numbers each appear once in 6 cases, 7 appears once and even numbers appear twice in 3 cases, and 7 appears twice and even numbers appear once in 3 cases. Let $A=$ "the product of the 3 points obtained from 3 rolls is divisible by 14", then $P(A)=6 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{2}+2 \times \frac{1}{6} \times\left(\frac{1}{2}\right)^{2}+3 \times\left(\frac{1}{6}\right)^{2} \times \frac{1}{2}=\frac{1}{3}$