Find the value of $a^4+b^4+c^4$ The problem:

The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$.  What is the sum of their fourth powers?

Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{align}
I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$, 
\begin{align}
p & = \alpha+\beta+\gamma \\
q & = \alpha \beta+\alpha\gamma+\beta\gamma \\
r & = \alpha\beta\gamma
\end{align}
Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:
\begin{align}
2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\
& = [(a+b+c)^2-(a^2+b^2+c^2)] \\
& = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\
& = 2ab+2ac+2bc
\end{align}
\begin{align}
q & = ab+ab+bc \\
& = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\
& = \frac{1}{2}[6^2-8] \\
& = 14
\end{align}
So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.
\begin{align}
f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\
f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\
f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\
\end{align}
\begin{align}
0 & = f(\alpha) + f(\beta) + f(\gamma) \\
& = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\
& = 5-6(8)+14(6)-3r\\
3r& = 41\\
r & = \frac{41}{3} \\
\end{align}
Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.
I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?

Euler's method works as follows. We put
$$
\begin{split}
a_1 &= a\\
a_2 &= b\\
a_3 &= c
\end{split}
$$
and
$$S_r = \sum_{j=1}^3 a_j^r$$
Consider the function:
$$f(u) = \sum_{j=1}^3\log\left(1-\frac{a_j}{u}\right)\tag{1}$$
Expanding this around infinity yields the series:
$$f(u) = -\sum_{k=1}^{\infty}\frac{S_k}{k u^k}\tag{2}$$
From (1) it is clear that $\exp\left[f(u)\right]$ is a third degree polynomial in $u^{-1}$. So, if we exponentiate the series (2), then terms of order 4 and higher in $u^{-1}$ will vanish. We can therefore compute $S_4$ by equating the fourth order term equal to zero. It's then convenient to write the exponential in factored form:
$$\exp\left[f(u)\right] = \exp\left(-\frac{S_1}{u}\right) \exp\left(-\frac{S_2}{2 u^2}\right) \exp\left(-\frac{S_3}{3u^3}\right) \exp\left(-\frac{S_4}{4u^4}\right) +\mathcal{O}(u^{-5})\tag{3}$$
Setting the coefficient of $u^{-4}$ equal to zero yields::
$$\frac{S_1^4}{24} - \frac{S_1^2 S_2}{4} + \frac{S_1 S_3}{3}  +\frac{S_2^2}{8} - \frac{S_4}{4} = 0 $$
So, we have:
$$S_4 = \frac{S_1^4}{6} - S_1^2 S_2 + \frac{4}{3} S_1 S_3 +\frac{S_2^2}{2}$$
which equals zero for the case at hand. To compute the sum over higher powers, we can derive a recursion relation using that the third degree polynomial $u^3\exp\left[f(u)\right]$ has zeroes at $u = a_j$. This is therefore also the characteristic polynomial for the recursion relation for $S_n$. 
Expanding (3) to third order yields:
$$u^3\exp\left[f(u)\right] = u^3 - S_1 u^2 + \left(\frac{S_1^2-S_2}{2}\right)u - \frac{S_1^3}{6} + \frac{S_1 S_2}{2}-\frac{S_3}{3}$$
This implies that:
$$S_{n+3} = S_1 S_{n+2} + \left(\frac{S_2 - S_1^2}{2}\right) S_{n+1} + \left(\frac{S_1^3}{6} - \frac{S_1 S_2}{2}+\frac{S_3}{3}\right) S_n$$
