parts of cubic equation

Question

Below there is a solved question from a book .

I could not understand how they got $S_{n+3}$ = $S_{n+1}$ - $S_n$

Answer 1

Since $\alpha$ is a root of the equation, $\alpha^3 - \alpha + 1 = 0$ that is $\alpha^3 = \alpha - 1$. Now, $\alpha^{n+3} = \alpha^3 \alpha^n = (\alpha - 1)\alpha^n = \alpha^{n+1} - \alpha^n$. Summing over all three roots gives the result.

Answer 2

All the exercise is centered onto Newton's Identities.
In particular, concerning your question, we have that putting $$ x^3 - x + 1 = e_0 x^3 - e_1 x^2 + e_2 x - e_3 = 0 $$ then $$ \begin{gathered} e_0 = 1 \hfill \\ e_1 = 0 = S_1 \hfill \\ e_2 = - 1 = \frac{1} {2}\left( {e_1 S_1 - S_2 } \right) = - \frac{1} {2}S_2 \hfill \\ e_3 = - 1 = \frac{1} {3}\left( {e_2 S_1 - e_1 S_2 + S_3 } \right) = \frac{1} {3}\left( { + 2 + S_3 } \right) \hfill \\ e_4 = 0 = \frac{1} {4}\left( {e_3 S_1 - e_2 S_2 + e_1 S_3 - S_4 } \right) = \frac{1} {4}\left( {e_3 S_1 - e_2 S_2 + e_1 S_3 - S_4 } \right) \hfill \\ \vdots \hfill \\ e_{n + 3} = 0 = \frac{{\left( { - 1} \right)^{n + 3} }} {{n + 3}}\left( {e_3 S_n - e_2 S_{n + 1} + e_1 S_{n + 2} - S_{n + 3} } \right) = \frac{{\left( { - 1} \right)^{n + 3} }} {{n + 3}}\left( { - S_n + S_{n + 1} - S_{n + 3} } \right) \hfill \\ \end{gathered} $$