Solve this question

Question : $$\begin{align}a+b+c &= 0\\ a^3 +b^3 +c^3 &= 12\\ a^5 +b^5 +c^5 &= 40\end{align}$$ Then , $a^4 + b^4 + c^4 = ?$

My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think $3abc = 12$ and got $abc = 4$ After that I am unable to link it with any other equation

Denote $S_n = a^n + b^n + c^n$. So we have $S_0 = 3, S_1 = 0, S_3 = 12$ and $S_5 = 40$.

Consider $f(x) = (x-a)(x-b)(x-c)$. Write $f(x) = x^3 - ux^2 - vx - w$. Then it is easy to see $u = 0$ and $v = -(ab + bc + ca) = \frac{S_2}2$. This means $a^3 = va + w$, so $a^n = va^{n-2} + wa^{n-3}$, and do the same on $b, c$ and summing together gives you $S_n = vS_{n-2} + wS_{n-3}$.

This should give you enough information set up a system of equations to see $S_4 = 8$. For instance, the first thing you can do is

$S_3 = vS_1 + wS_0$, so $12 = 3w$ and $w = 4$.

Hint: $$(a+b+c)^4=a^4+b^4+c^4+4a^3(b+c)+4b^3(a+c)+4c^3(a+b)+6(a^2b^2+b^2c^2+c^2a^2)+12abc(a+b+c)=0$$