<s>[INST] What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If 
$$ \alpha + \beta + \gamma = 14  $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?

Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \gamma^{2}  + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196$$
$$\alpha^{2} + \beta^{2} + \gamma^{2}  + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 \implies 84 + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196  $$
$$ \alpha \beta + \beta \gamma + \alpha \gamma = 56 $$
Also
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{3} + \beta^{3} + \gamma^{3}  + 3(\alpha^{2} \beta + \alpha^{2} \gamma + \beta^{2} \alpha + \beta^{2} \gamma + \gamma^{2} \alpha + \gamma^{2} \beta) + 3 \alpha \beta \gamma = 2744$$
$$ 584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + 6 \alpha \beta \gamma$$
$$ \alpha \beta \gamma = 32$$
So that
$$ (\alpha + \beta + \gamma)^{4} = ( \alpha^{2} + \beta^{2} + \gamma^{2}  + 2(\alpha \beta + \beta \gamma + \alpha \gamma) )^{2} $$
$$ 38416 = (\alpha^{2} + \beta^{2} + \gamma^{2})^{2} + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2}) + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)^{2}$$
$$ 38416 = \left[ \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta)^{2} + (\beta \gamma)^{2} + (\alpha \gamma)^{2} \right) \right] + 4 (56)(84) + 4 (56^{2}) $$
$$ 7056 = \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta + \alpha \gamma + \beta \gamma)^{2} - 2\alpha \beta \gamma(\alpha + \beta + \gamma ) \right) $$
$$ 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(32)(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} = 2576$$
 [/INST] Define $A,B,C$ such that $(x-\alpha)(x-\beta)(x-\gamma) = x^3 - Ax^2 + Bx - C$.
For any $n \ge 1$, let $p_n = \alpha^n + \beta^n + \gamma^n$. In terms of $p_n$, the question becomes:

Given $p_1 = 14, p_2 = 84, p_3 = 584$, what is $p_4$?

There is a bunch of Newton's identities connecting $p_n$ and the coefficients $A,B,C$ above.
Using these identities, the computation of $p_4 = \alpha^4 + \beta^4 + \gamma^4$ is mechanical.


*

*$p_1 = A \\ \quad \implies A = 14$ 

*$p_2 = A p_1 - 2B  \\ \quad  \implies  B = \frac12(A^2 - p_2) = \frac12(14^2 - 84) = 56$

*$p_3 = A p_2 - B p_1 + 3C \\ \quad \implies C = \frac13 ( p_3 - Ap_2 + B p_1)
= \frac13(584 - 14\cdot 84 + 56\cdot 14) = 64$
Finally, we have
$$p^4 = Ap_3 - Bp_2 + Cp_1 = 14\cdot 584 - 56\cdot 84 + 64\cdot 14 = 4368$$
This is different from what you have got. 
When you compute $\alpha\beta\gamma$, you have make a mistake in one of the coefficient. The correct step should be
$$584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + \color{red}{3} \alpha \beta \gamma\\
\implies \alpha\beta\gamma = \color{red}{64}
$$
Other than this, the rest of your derivation seems fine. In fact, if you substitute $32$ by $64$ in last two lines, you get the right answer.
$$\require{cancel}
 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(\color{red}{\cancelto{64}{\color{gray}{32}}})(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} = 
\color{red}{\cancelto{4368}{\color{gray}{2576}}}
$$
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