Q6 polynomial complex Find the equation whose roots are the fourth powers of the roots of the equation $x^3 + x + 1 = 0$. Hence find the sum of the fourth powers of the roots of the equation  $x^3 + x + 1 = 0$
I am really stuck. The answer at the back says cubic is $u^3 - 2u^2 + 5u -1 = 0$ and sum of roots is 2

So we seek a cubic in $y = x^4, $ where $x^3+x+1=0$.  
Hence we have $y = x^4 = x(x^3) = -x(x+1) = -x^2-x$, using which, $y^2 = (x^4)^2 = x^2(x+1)^2 = x^4+2x^3+x^2 = -3x-2$.
Similarly, $y^3 = (x^4)^3 = -x^3(x+1)^3 = 5x^2-x-3$.  
Now it is easy to eliminate (the powers of) $x$ and get $y^3-2y^2 +5y = 1\dots$
