Let $a$, $b$, $c$, $d$ be the roots of $x^4 + x + 1 = 0$. Find $a^4 + b^4 + c^4 + d^4$. 
Let $p(x) = x^4 + x + 1 = 0$, and let $a$, $b$, $c$, $d$ be its roots.
  Find $a^4 + b^4 + c^4 + d^4$.

I have no idea how to start solving this problem.

As an alternative, let consider
$$(x-a)(x-b)(x-c)(x-d)=$$
$$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$
then


*

*$S_1=a+b+c+d=0$

*$S_2=ab+ac+ad+bc+bd+cd=0$

*$S_3=abc+abd+acd+bcd=-1$

*$S_4=abcd=1$


and by Newton's sums we have that


*

*$P_1=a+b+c+d=S_1=0$

*$P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2=0$

*$P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3=-3$

*$P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4=-4$

