What's $f(4)$ ?

Relevant wiki: Newton's Identities

$a+b+c = 0\\ a^2+b^2+c^2=p\\ a^3+b^3+c^3=3\\ a^4+b^4+c^4=q\\ a^5+b^5+c^5=10$

Let $ab+ac+bc = r$ and $abc = s$ .

Use Newton's sums to find the values of $p,q,r$ and $s$ .

$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+ac+bc)$

$\implies p=-2r$ - Eq. (1)

$a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2) - (ab+ac+bc)(a+b+c) + 3abc\\ 3 = 3s\\ s=1 \implies abc=1$

$a^4+b^4+c^4 = (a+b+c)(a^3+b^3+c^3) - (ab+ac+bc)(a^2+b^2+c^2) + abc(a+b+c)$

$\implies q=-pr$ - Eq. (2)

$a^5+b^5+c^5 = (a+b+c)(a^4+b^4+c^4) - (ab+ac+bc)(a^3+b^3+c^3) + abc(a^2+b^2+c^2)$

$\implies 10 = -3r + p$ - Eq. (3)

We now have 3 linear equations with 3 variables, and we can proceed to solve them.

Substitute Eq. (1) into Eq. (3):

$10 = -3r - 2r\\ r = -2 \implies ab+ac+bc = -2$

Therefore,

$p= -2(-2) = 4 \implies a^2+b^2+c^2 = 4$

And finally,

$q=-pr = -(4)(-2) = 8 \implies a^4+b^4+c^4 = \boxed{8}$