<s>[INST] 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?
 [/INST] Making the given substitution yields $y^{4/3}=y+1$, or $y^4=(y+1)^3$ or $y^4-y^3-3y^2-3y-1=0$. The roots of this polynomial are $\alpha^3,\beta^3,\gamma^3,\delta^3$, so we can work out
$$\alpha^6+\beta^6+\gamma^6+\delta^6=(\alpha^3+\beta^3+\gamma^3+\delta^3)^2-2(\alpha^3\beta^3+\cdots)=1^2-2(-3)=7$$
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