Answer: 6
If we write the roots of \(x^3+x-1=0\) as \(\alpha,\beta,\gamma\), then by Vieta’s relations
\(\alpha+\beta+\gamma=0\), \(\alpha\beta+\beta\gamma+\gamma\alpha=1\), and \(\alpha\beta\gamma=1\).
Now let \(y=x^2\). Then \(x=\sqrt{y}\), so from \(x^3+x-1=0\) we get
\(x(x^2+1)=1\).
Since \(x^2=y\), this becomes
\(x(y+1)=1\), so after squaring and using \(x^2=y\),
\(y(y+1)^2=1\).
Expanding gives
\(y^3+2y^2+y-1=0\).
Therefore \(\alpha^2,\beta^2,\gamma^2\) are roots of \(y^3+2y^2+y-1=0\).
(i) \(S_2=\alpha^2+\beta^2+\gamma^2\).
Using \((\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)\),
\(0=S_2+2\cdot 1\), so \(S_2=-2\).
Also
\(S_4=\alpha^4+\beta^4+\gamma^4=(\alpha^2+\beta^2+\gamma^2)^2-2(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)\).
Now \(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)=1^2-0=1\).
Hence
\(S_4=(-2)^2-2(1)=4-2=2\).
(ii) Since \(\alpha,\beta,\gamma\) satisfy \(r^3=1-r\), multiplying by \(r^{n-3}\) shows \(r^n=r^{n-3}-r^{n-2}\) for each root \(r\), and therefore
\(S_n=S_{n-3}-S_{n-2}\).
So
\(S_6=S_3-S_4\), and from the cubic \(r^3=1-r\), summing over the roots gives
\(S_3=3-S_1=3-0=3\).
Thus \(S_6=3-2=1\).
Then
\(S_8=S_5-S_6\).
Also \(S_5=S_2-S_3=-2-3=-5\), so
\(S_8=-5-1=-6\).
Final answers: \(S_2=-2\), \(S_4=2\), \(S_6=1\), \(S_8=-6\).
Let the roots of \(x^3+x-1=0\) be \(\alpha,\beta,\gamma\). Then by Vieta’s formulae
\(\alpha+\beta+\gamma=0\), \(\alpha\beta+\beta\gamma+\gamma\alpha=1\), \(\alpha\beta\gamma=1\).
To obtain an equation for the squares of the roots, set \(y=x^2\), so \(x=\sqrt{y}\). The given equation becomes
\(x^3+x-1=0\Rightarrow x(x^2+1)=1\).
Since \(x^2=y\), this is
\(x(y+1)=1\).
Squaring gives
\(x^2(y+1)^2=1\).
Now \(x^2=y\), so
\(y(y+1)^2=1\).
Expanding and rearranging:
\(y(y^2+2y+1)=1\)
\(y^3+2y^2+y-1=0\).
So \(\alpha^2,\beta^2,\gamma^2\) are roots of \(y^3+2y^2+y-1=0\).
(i) Define \(S_n=\alpha^n+\beta^n+\gamma^n\).
First,
\(S_2=\alpha^2+\beta^2+\gamma^2\).
Using \((\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)\), we get
\(0=S_2+2(1)\), hence
\(S_2=-2\).
Now find \(S_4\):
\(S_4=\alpha^4+\beta^4+\gamma^4=(\alpha^2+\beta^2+\gamma^2)^2-2(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)\).
Also
\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)\).
Substituting the known values gives
\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=1^2-2\cdot 1\cdot 0=1\).
Therefore
\(S_4=(-2)^2-2(1)=4-2=2\).
(ii) For each root \(r\in\{\alpha,\beta,\gamma\}\), the cubic gives \(r^3=1-r\). Multiplying by \(r^{n-3}\) yields
\(r^n=r^{n-3}-r^{n-2}\).
Adding over \(r=\alpha,\beta,\gamma\), we obtain the recurrence
\(S_n=S_{n-3}-S_{n-2}\).
Now \(S_3=\alpha^3+\beta^3+\gamma^3\). Since \(r^3=1-r\), summing gives
\(S_3=3-(\alpha+\beta+\gamma)=3-0=3\).
Hence
\(S_6=S_3-S_4=3-2=1\).
Next,
\(S_5=S_2-S_3=-2-3=-5\), so
\(S_8=S_5-S_6=-5-1=-6\).
Therefore \(S_6=1\) and \(S_8=-6\).
