<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] Let $x=\alpha+\beta+\gamma$, $y=\alpha^2+\beta^2+\gamma^2$ and $z=\alpha^3+\beta^3+\gamma^3$ for any real numbers $\alpha,\beta,\gamma$. Then we have the identity:
$$\alpha^4+\beta^4+\gamma^4=\frac16\left(x^4-6x^2y+3y^2+8xz\right).\tag{1}$$
You can verify this by expanding everything. In our case, $x=14,y=84$ and $z=584$, so just substitute in the numbers to solve the problem. I get the answer of $4368$.
(Note: I came up with the expression $(1)$ with a computer running Mathematica. It might not be very viable in a competition setting.)
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