Answer: (i) The roots \(\alpha,\beta,\gamma\) satisfy
\(\alpha+\beta+\gamma=-2,\quad \alpha\beta+\beta\gamma+\gamma\alpha=1,\quad \alpha\beta\gamma=-7\)
by Vieta’s formulae for \(x^{3}+2x^{2}+x+7=0\).
Now let
\(y=\dfrac{\alpha}{\beta\gamma}\).
Since \(\beta\gamma=\dfrac{-7}{\alpha}\), we get \(y=\dfrac{\alpha^{2}}{-7}\), so \(\alpha^{2}=-7y\). Thus each root \(\alpha,\beta,\gamma\) gives a corresponding value of \(y\), and the resulting cubic in \(y\) has roots \(\dfrac{\alpha}{\beta\gamma},\dfrac{\beta}{\gamma\alpha},\dfrac{\gamma}{\alpha\beta}\).
Using the standard transformation for the reciprocal-type quantities, these three values are the roots of \(49y^{3}+14y^{2}-27y+7=0\).
(ii) Let
\(u=\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}},\quad v=\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}},\quad w=\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}.\)
Then
\(u=\left(\dfrac{\alpha}{\beta\gamma}\right)^{2},\ v=\left(\dfrac{\beta}{\gamma\alpha}\right)^{2},\ w=\left(\dfrac{\gamma}{\alpha\beta}\right)^{2}.\)
So \(u,v,w\) are the squares of the roots from part (i). If those roots are \(r_1,r_2,r_3\), then from \(49y^{3}+14y^{2}-27y+7=0\),
\(r_1+r_2+r_3=-\dfrac{14}{49}=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=\dfrac{-27}{49}.\)
Therefore
\(u+v+w=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)\)
\(=\left(-\dfrac{2}{7}\right)^2-2\left(-\dfrac{27}{49}\right)=\dfrac{4}{49}+\dfrac{54}{49}=\dfrac{58}{49}.\)
(iii) Let
\(S_3=\dfrac{\alpha^{3}}{\beta^{3}\gamma^{3}}+\dfrac{\beta^{3}}{\gamma^{3}\alpha^{3}}+\dfrac{\gamma^{3}}{\alpha^{3}\beta^{3}}.\)
Using \(u,v,w\) as above, we have \(S_3=u^{3/2}+v^{3/2}+w^{3/2}\), but it is easier to use the roots \(r_1,r_2,r_3\) from part (i):
\(S_3=r_1^{3}+r_2^{3}+r_3^{3}.\)
From the cubic in part (i), the elementary symmetric sums are
\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49},\quad r_1r_2r_3=-\dfrac{1}{7}.\)
Now use
\(r_1^{3}+r_2^{3}+r_3^{3}=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)
So
\(S_3=\left(-\dfrac{2}{7}\right)^3-3\left(-\dfrac{2}{7}\right)\left(-\dfrac{27}{49}\right)+3\left(-\dfrac{1}{7}\right)\)
\(=-\dfrac{8}{343}-\dfrac{162}{343}-\dfrac{147}{343}=-\dfrac{317}{343}.\)
Hence the exact value is \(\boxed{-\dfrac{317}{343}}\).
Let the roots of \(x^{3}+2x^{2}+x+7=0\) be \(\alpha,\beta,\gamma\). By Vieta’s formulae,
\(\alpha+\beta+\gamma=-2,\quad \alpha\beta+\beta\gamma+\gamma\alpha=1,\quad \alpha\beta\gamma=-7.\)
(i) We want the cubic satisfied by the three numbers
\(\dfrac{\alpha}{\beta\gamma},\quad \dfrac{\beta}{\gamma\alpha},\quad \dfrac{\gamma}{\alpha\beta}.\)
Set
\(u=\dfrac{\alpha}{\beta\gamma}.\)
Since \(\beta\gamma=\dfrac{\alpha\beta\gamma}{\alpha}=\dfrac{-7}{\alpha}\), we have
\(u=\dfrac{\alpha}{-7/\alpha}=-\dfrac{\alpha^{2}}{7}.\)
So \(\alpha^{2}=-7u\), and similarly \(\beta^{2}=-7v\), \(\gamma^{2}=-7w\), where
\(v=\dfrac{\beta}{\gamma\alpha},\quad w=\dfrac{\gamma}{\alpha\beta}.\)
Now compute the symmetric sums of \(u,v,w\).
First,
\(u+v+w=-\dfrac{1}{7}(\alpha^{2}+\beta^{2}+\gamma^{2}).\)
Also
\(\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=(-2)^2-2(1)=2.\)
Hence
\(u+v+w=-\dfrac{2}{7}.\)
Next,
\(uv+vw+wu=\dfrac{1}{49}(\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2})\).
Now
\((\alpha\beta+\beta\gamma+\gamma\alpha)^2=\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+2\alpha\beta\gamma(\alpha+\beta+\gamma),\)
so
\(1=\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+2(-7)(-2)\).
Therefore
\(\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}=1-28=-27,\)
and
\(uv+vw+wu=-\dfrac{27}{49}.\)
Finally,
\(uvw=\dfrac{\alpha\beta\gamma}{\alpha^{2}\beta^{2}\gamma^{2}}=\dfrac{1}{\alpha\beta\gamma}=-\dfrac{1}{7}.\)
Therefore the monic cubic with roots \(u,v,w\) is
\(t^{3}-(u+v+w)t^{2}+(uv+vw+wu)t-uvw=0,\)
that is
\(t^{3}+\dfrac{2}{7}t^{2}-\dfrac{27}{49}t+\dfrac{1}{7}=0.\)
Multiplying by 49 gives
\(49t^{3}+14t^{2}-27t+7=0,\)
so the stated numbers are indeed roots.
(ii) Let
\(u=\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}},\quad v=\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}},\quad w=\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}.\)
Then \(u,v,w\) are the roots from part (i) squared, so if we denote those roots by \(r_1,r_2,r_3\), then \(u=r_1^2\), \(v=r_2^2\), \(w=r_3^2\).
We use the identity
\(u+v+w=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).\)
From part (i),
\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49}.\)
Hence
\(u+v+w=\left(-\dfrac{2}{7}\right)^2-2\left(-\dfrac{27}{49}\right)=\dfrac{4}{49}+\dfrac{54}{49}=\dfrac{58}{49}.\)
So
\(\dfrac{\alpha^{2}}{\beta^{2}\gamma^{2}}+\dfrac{\beta^{2}}{\gamma^{2}\alpha^{2}}+\dfrac{\gamma^{2}}{\alpha^{2}\beta^{2}}=\boxed{\dfrac{58}{49}}.\)
(iii) We want
\(\dfrac{\alpha^{3}}{\beta^{3}\gamma^{3}}+\dfrac{\beta^{3}}{\gamma^{3}\alpha^{3}}+\dfrac{\gamma^{3}}{\alpha^{3}\beta^{3}}.\)
Using the same roots \(r_1,r_2,r_3\) from part (i), this is
\(r_1^{3}+r_2^{3}+r_3^{3}.\)
Now apply
\(r_1^{3}+r_2^{3}+r_3^{3}=(r_1+r_2+r_3)^3-3(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)+3r_1r_2r_3.\)
Substituting the sums from part (i),
\(r_1+r_2+r_3=-\dfrac{2}{7},\quad r_1r_2+r_2r_3+r_3r_1=-\dfrac{27}{49},\quad r_1r_2r_3=-\dfrac{1}{7},\)
gives
\(r_1^{3}+r_2^{3}+r_3^{3}=\left(-\dfrac{2}{7}\right)^3-3\left(-\dfrac{2}{7}\right)\left(-\dfrac{27}{49}\right)+3\left(-\dfrac{1}{7}\right).\)
So
\(r_1^{3}+r_2^{3}+r_3^{3}=-\dfrac{8}{343}-\dfrac{162}{343}-\dfrac{147}{343}=-\dfrac{317}{343}.\)
Therefore the exact value is \(\boxed{-\dfrac{317}{343}}\).
