The equation $x^4-x^3-1=0$ has roots $\alpha, \beta, \gamma, \delta$. Find the equations with roots $\alpha^6, \beta^6, \gamma^6, \delta^6$. I was able to do this using the substitution $y=x^3$. I obtained the equation $y^4-y^3-3y^2-3y-1=0$ and then applied applied the formula $\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta$ to get $S_6$. The mark schemes mentions an alternative way way which I do not understand.
Here it is. 'Use $S_{n+4}=S_n + S_{n+3}$ ' and then goes on to compute $S_{-1}, S_2, S_3, S_4, S_5$ and finally $S_6$ How did they get $S_{n+4}=S_n + S_{n+3}$ in the first place?
Could someone please explain?
Note that $\alpha^4-\alpha^3-1=0$. If you multiply this by $\alpha^n$ and add the corresponding results for $\beta, \gamma, \delta$ you get what you are looking for.
Note that it is useful to know that $S_0=4$
It is easy to add spurious solutions here: what you are looking for is $4$ solutions and therefore a quartic equation.