Complex

If $\sqrt{1+i}$ is a root of this particular polynomial, then its negative must also be a root. That gives us an equation

$f(x) = x^2 - (1+i)$

however, this still contains a complex part. Thus this needs another factor that will cancel this out. This gives us

$f(x) = [x^2 - (1+i)][x^2 - (1-i)]$

$f(x) = x^4 - 2x^2 + 2$

From here, we use Newton's sums to find the sum cubes of the roots.

$P_1 = 0$

$P_2 = 0 - 2(-2) = 4$

$P_3 = 0(P_2) - 2P_1 + 3(0) = \boxed {0}$