Newton's fifth degree dilemma

$x^{3} + 3x + 3 = 0$
$\displaystyle \sum_{cyclic} \alpha = 0 , \sum_{cyclic} \alpha\beta = 3$
$x^{3} = -3x - 3$
$x^{5} = -3x^{3} - 3x^{2}$
$x^{5} = -3(-3x-3) - 3x^{2} = 9x + 9 - 3x^{2}$
$\displaystyle \sum_{cyclic} \alpha^{5} = 9\displaystyle \sum_{cyclic}\alpha+ 9 \sum_{cyclic} 1 - 3\sum_{cyclic} \alpha^{2}$
$S = 9(0) + 9(3) - 3(0^{2} -2(3)) = 27 + 18 = 45$

For easy typing purposes, let $\alpha = a,\;\beta=b$ and $\gamma=c$

$x^3+3x+3=0$

From Vieta's formula, we know that:

$a+b+c=0\\ ab+ac+bc=3\\ abc=-3$

Use Newton's sums to find the value of $a^5+b^5+c^5$ :

$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+ac+bc) = 0^2 - 2(3) = -6\\ a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2) -(ab+ac+bc)(a+b+c)+3abc = 0(-6)-3(0)+3(-3) = -9\\ a^4+b^4+c^4 = (a+b+c)(a^3+b^3+c^3) - (ab+ac+bc)(a^2+b^2+c^2) + abc(a+b+c) = 0(-9) - 3(-6)+(-3)(0) = 18\\ a^5+b^5+c^5 = (a+b+c)(a^4+b^4+c^4) - (ab+ac+bc)(a^3+b^3+c^3) +abc(a^2+b^2+c^2) = 0(27) - 3(-9)+(-3)(-6) = 27+18 = 45$

$\implies \alpha^5+\beta^5+\gamma^5 =\boxed{45}$