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?

"Further mathematics" sounds like IB. So, they may suggest a solution using linear recursions.
The given polynomial
$$x^4-\color{green}{1}\cdot x^3 +\color{blue}{0}\cdot x^2 +\color{blue}{0}\cdot x - 1$$ belongs to the linear recursion
$$x_{n+4} = x_{n+3}+x_n$$
with (using Vieta)

*

*$x_0 = \alpha^0 + \beta^0 + \gamma^0 + \delta^0 = 4$

*$x_1 = \alpha + \beta + \gamma + \delta =\color{green}{1}$

*$x_2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2=(\alpha + \beta + \gamma + \delta)^2 -2\cdot \color{blue}{0} = 1$

*$x_3 = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 =(\alpha + \beta + \gamma + \delta)^3-3(\alpha + \beta + \gamma + \delta)\cdot\color{blue}{0}-3\cdot\color{blue}{0} =1$
Now, since
$$\alpha^6 + \beta^6 + \gamma^6 + \delta^6 = x_6$$
you just use the recursion
$$x_4= x_3+x_0 = 1+4 = 5$$
$$x_5= x_4+x_1 = 5+1 = 6$$
$$\boxed{x_6}= x_5+x_2 = 6+1 = \boxed{7}$$
