parts of cubic equation Below there is a solved question from  a  book . 
I could not understand how they got $S_{n+3}$ = $S_{n+1}$ - $S_n$
 A: 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.
A: 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
$$
\begin{gathered}
  e_0  = 1 \hfill \\
  e_1  = 0 = S_1  \hfill \\
  e_2  =  - 1 = \frac{1}
{2}\left( {e_1 S_1  - S_2 } \right) =  - \frac{1}
{2}S_2  \hfill \\
  e_3  =  - 1 = \frac{1}
{3}\left( {e_2 S_1  - e_1 S_2  + S_3 } \right) = \frac{1}
{3}\left( { + 2 + S_3 } \right) \hfill \\
  e_4  = 0 = \frac{1}
{4}\left( {e_3 S_1  - e_2 S_2  + e_1 S_3  - S_4 } \right) = \frac{1}
{4}\left( {e_3 S_1  - e_2 S_2  + e_1 S_3  - S_4 } \right) \hfill \\
   \vdots  \hfill \\
  e_{n + 3}  = 0 = \frac{{\left( { - 1} \right)^{n + 3} }}
{{n + 3}}\left( {e_3 S_n  - e_2 S_{n + 1}  + e_1 S_{n + 2}  - S_{n + 3} } \right) = \frac{{\left( { - 1} \right)^{n + 3} }}
{{n + 3}}\left( { - S_n  + S_{n + 1}  - S_{n + 3} } \right) \hfill \\ 
\end{gathered} 
$$
