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?

Define the polynomial function $p$ by
$$p(x) = x^4-x^3-1$$
Also define the function $q$ by
$q(x)=p\left(x^{1/3}\right)$. Then,
$$q(t)=t^{4/3}-t-1$$
By definition of $q$, we have that $\alpha^3,\beta^3,\gamma^3,\delta^3$ are roots of the equation $q(t)=0$.
$$q(t)=0\iff t^4 - (t+1)^3=0$$
The polynomial $t^4-(t+1)^3$ is a fourth degree polynomial in $t$, and we know four of its roots -- and since any polynomial of degree $n\geqslant 1$ has exactly $n$ roots, these are the $\textit{only}$ roots.
Now, Vieta's formulae give
$$\alpha^3+\beta^3+\gamma^3+\delta^3 = 1 \:\:\text{ and } \:\:\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3=-3$$
Thus, $$\begin{align}\alpha^6+\beta^6+\gamma^6+\delta^6&=\left(\alpha^3+\beta^3+\gamma^3+\delta^3\right)^2-2\left(\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3\right)\\ &=7\end{align}$$
The problem is solved. $\square$
