The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?

Let $P_n:=\alpha^n+\beta^n+\gamma^n+\delta^n$. Then, Newton's Sums tell us that $$1\cdot P_1-1=0\iff P_1=1\\1\cdot P_2-1\cdot P_1+2\cdot0=0\iff P_2=1\\1\cdot P_3-1\cdot P_2+0\cdot P_1+3\cdot 0=0\iff P_3=1\\\vdots$$
