Let us consider a cubic polynomial whose roots are $a$, $b$, $c$. We get $P(x) = x^3 - px^2 + qx - r$, where $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$. We observe that
$$\frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-c)(b-a)} + \frac{c^n}{(c-a)(c-b)} = \sum_{\text{cyclic}} \frac{-a^n(b-c)}{(a-b)(b-c)(c-a)}$$
Let us write
$$S_n = \sum_{\text{cyclic}} \frac{a^n (b-c)}{-(a-b)(b-c)(c-a)}$$
It is easy to see that $S_1 = 0$ and $S_2 = 1$. Since $a$, $b$, $c$ are the roots of $P(x) = 0$, we have $a^3 - pa^2 + qa - r = 0$, $b^3 - pb^2 + qb - r = 0$, $c^3 - pc^2 + qc - r = 0$. Multiply the first by $b-c$, the second by $c-a$ and the third by $a-b$, and adding all these and dividing the sum by $-(a-b)(b-c)(c-a)$, we obtain $S_3 - pS_2 + qS_1 = 0$. Hence $S_3 = p$. Now multiply the first by $a$, the second by $b$ and the third by $c$ and divide throughout by $-(a-b)(b-c)(c-a)$ to get $S_4 - pS_3 + qS_2 - rS_1 = 0$. Hence $S_4 = p^2 - q$. Similarly, $S_5 = p(p^2 - q) - qp + r = p^3 - 2pq + r$. We also get $S_6 - pS_5 + qS_4 - rS_3 = 0$. This gives
$$S_6 = p(p^3 - 2pq + r) - q(p^2 - q) + rp = p^4 - 3p^2q + 2pr + q^2.$$
We can write it as $S_6 = p^2(p^2 - 3q) + 2pr + q^2$. But $p^2 - 3q = (a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca > 0$. Hence
$$S_6 > 2pr + q^2 = 2abc(a + b + c) + (ab + bc + ca)^2 \geq 6 + 9 = 15.$$