An algebra problem by Akshat Sharda

$\begin{aligned} x^4+4x+4 & = 0 \\ \Rightarrow x^4 & = -4(x+1) \\ x^{12} & = -4^3(x+1)^3 \\ & = -64(x^3+3x^2+3x+1) \\ \Rightarrow \alpha^{12} + \beta^{12} + \gamma^{12} + \delta^{12} & = -64\left(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 +3(\alpha^2 + \beta^2 + \gamma^2 + \delta^2)+3(\alpha + \beta + \gamma + \delta) + 4\right) \end{aligned}$

By Vieta's formulas, we have:

$\begin{aligned}\alpha + \beta + \gamma + \delta & = 0 \\ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 & = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta) \\ & = 0^2 -2(0) = 0 \\ \alpha^3 + \beta^3 + \gamma^3 + \delta^3 & = (\alpha + \beta + \gamma + \delta)(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) \\ & \quad - (\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)(\alpha + \beta + \gamma + \delta) + 3(\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta) \\ & = (0)(0) -(0)(0) +3(-4) = -12 \end{aligned}$

Therefore,

$\begin{aligned} \alpha^{12} + \beta^{12} + \gamma^{12} + \delta^{12} & = -64\left(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 +3(\alpha^2 + \beta^2 + \gamma^2 + \delta^2)+3(\alpha + \beta + \gamma + \delta) + 4\right) \\ & = -64 (-12+3(0)+3(0)+4) = -64(-8) = 2^9 \end{aligned}$

$\Rightarrow \log_2 \left(\alpha^{12} + \beta^{12} + \gamma^{12} + \delta^{12}\right) = \boxed{9}$