Find the value of $a^4+b^4+c^4$

$f(\alpha)=0$ implies that $\alpha f(\alpha)=0$.

So, $\alpha^4-6\alpha^3+14\alpha^2-\frac{41}{3}\alpha=0$

We have similar results for $\beta$ and $\gamma$.

So, $\alpha^4+\beta^4+\gamma^4=6(\alpha^3+\beta^3+\gamma^3)-14(\alpha^2+\beta^2+\gamma^2)+\frac{41}{3}(\alpha+\beta+\gamma)$

Hint:

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\iff ab+bc+ca=?$$

$$a^3+b^3+c^3-3abc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}\iff abc=?$$

Now $$a^4+b^4+c^4=(\underbrace{a^2+b^2+c^2})^2-2(a^2b^2+b^2c^2+c^2a^2)$$

$$a^2b^2+b^2c^2+c^2a^2=(\underbrace{ab+bc+ca})^2-2\underbrace{abc}(\underbrace{a+b+c})$$