Let S(n) denote the sum of the nth powers of the roots. Then the polynomial implies S(n) = S(n-2)-S(n-3).

S(0)=3 is trivial.

S(1)=0 from the coefficient of x^2 being 0.

Let C be the cross-product of the roots (C=P*Q+P*R+Q*R), which makes C=-1 from the coefficient of x. Then S(2) = S(1)^2-2*C = 0^2 - 2*(-1) = 2.

The sequence S(n) starting at n=0 is 3, 0, 2, -3, 2, -5, 5, -7, 10, ...

S(8) = 10, the answer to the problem.