Let a, b, c be three roots of the equation $x^3+x^2-333x-1002=0$, then $(\displaystyle\sum (a^3)-2\displaystyle\sum a)$ equals to?

For x^3 + x^2 - 333x - 1002 = 0, the sum of roots a+b+c = -1. Using Newton's sums or the fact that the roots satisfy the equation, a^3 = -a^2 + 333a + 1002. Summing for a, b, c gives sum(a^3) = -sum(a^2) + 333*sum(a) + 3006. Since sum(a^2) = (sum a)^2 - 2*sum(ab) = (-1)^2 - 2(-333) = 667, we find sum(a^3) = -667 + 333(-1) + 3006 = 2006. Then 2006 - 2(-1) = 2008.