Answer: (i) Let \(y=\frac{1}{x}\), so \(x=\frac{1}{y}\). Substituting into \(x^{4}+x^{3}+cx^{2}+4x-2=0\) gives
\(\frac{1}{y^{4}}+\frac{1}{y^{3}}+\frac{c}{y^{2}}+\frac{4}{y}-2=0\).
Multiplying by \(y^{4}\) gives \(1+y+cy^{2}+4y^{3}-2y^{4}=0\), so an equation with roots \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma},\frac{1}{\delta}\) is
\(2y^{4}-4y^{3}-cy^{2}-y-1=0\).
(ii) For \(x^{4}+x^{3}+cx^{2}+4x-2=0\), the roots \(\alpha,\beta,\gamma,\delta\) satisfy
\(\alpha+\beta+\gamma+\delta=-1\), \(\sum \alpha\beta=c\), and \(\alpha\beta\gamma\delta=-2\).
Then
\(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}=(\alpha+\beta+\gamma+\delta)^{2}-2\sum \alpha\beta\)
\(= (-1)^{2}-2c=1-2c\).
Also, the roots of the reciprocal equation are \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma},\frac{1}{\delta}\), so its coefficient relations give
\(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}=2\),
and
\(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}+\frac{1}{\delta^{2}}=\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}\right)^{2}-2\sum \frac{1}{\alpha\beta}\).
Now \(\sum \frac{1}{\alpha\beta}=\frac{\alpha\beta+\alpha\gamma+\cdots}{\alpha\beta\gamma\delta}=\frac{c}{-2}=-\frac{c}{2}\), so
\(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}+\frac{1}{\delta^{2}}=2^{2}-2\left(-\frac{c}{2}\right)=4+c\).
(iii) Let
\(S=\left(\alpha-\frac{1}{\alpha}\right)^{2}+\left(\beta-\frac{1}{\beta}\right)^{2}+\left(\gamma-\frac{1}{\gamma}\right)^{2}+\left(\delta-\frac{1}{\delta}\right)^{2}\).
Expanding term by term,
\(S=(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2})+\left(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}+\frac{1}{\delta^{2}}\right)-2\left(\alpha\cdot\frac{1}{\alpha}+\beta\cdot\frac{1}{\beta}+\gamma\cdot\frac{1}{\gamma}+\delta\cdot\frac{1}{\delta}\right)\).
So
\(S=(1-2c)+(4+c)-8=-c-3\).
(iv) If \(c=-3\), then \(S=0\). Since each term \(\left(\alpha-\frac{1}{\alpha}\right)^{2}\) is a square of a real number, if all the roots were real then every term would have to be zero. Hence each root would satisfy \(\alpha=\pm 1\), and similarly for \(\beta,\gamma,\delta\).
But this is impossible, because the product of the roots is \(\alpha\beta\gamma\delta=-2\), whereas a product of four numbers each equal to \(\pm 1\) can only be \(\pm 1\). Therefore the roots are not all real.
(i) Put \(y=\frac{1}{x}\), so that \(x=\frac{1}{y}\). Then
\(\left(\frac{1}{y}\right)^4+\left(\frac{1}{y}\right)^3+c\left(\frac{1}{y}\right)^2+4\left(\frac{1}{y}\right)-2=0\).
Multiply by \(y^4\):
\(1+y+cy^2+4y^3-2y^4=0\).
Rearranging,
\(2y^4-4y^3-cy^2-y-1=0\).
This has roots \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma},\frac{1}{\delta}\).
(ii) For the quartic \(x^4+x^3+cx^2+4x-2=0\), Vieta's relations give
\(\alpha+\beta+\gamma+\delta=-1\), \(\sum \alpha\beta=c\), and \(\alpha\beta\gamma\delta=-2\).
Now
\(\alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2\sum \alpha\beta\)
\(= (-1)^2-2c=1-2c\).
For the reciprocals, the equation from part (i) is
\(2y^4-4y^3-cy^2-y-1=0\).
Dividing by 2 gives a monic quartic
\(y^4-2y^3-\frac{c}{2}y^2-\frac{1}{2}y-\frac{1}{2}=0\),
with roots \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma},\frac{1}{\delta}\). Hence
\(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}=2\),
and
\(\frac{1}{\alpha\beta}+\frac{1}{\alpha\gamma}+\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}+\frac{1}{\beta\delta}+\frac{1}{\gamma\delta}=-\frac{c}{2}\).
Therefore
\(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}+\frac{1}{\delta^2}=\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}\right)^2-2\sum \frac{1}{\alpha\beta}\)
\(=2^2-2\left(-\frac{c}{2}\right)=4+c\).
(iii) Let
\(S=\left(\alpha-\frac{1}{\alpha}\right)^2+\left(\beta-\frac{1}{\beta}\right)^2+\left(\gamma-\frac{1}{\gamma}\right)^2+\left(\delta-\frac{1}{\delta}\right)^2\).
Expanding each square gives
\(S=(\alpha^2+\beta^2+\gamma^2+\delta^2)+\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}+\frac{1}{\delta^2}\right)-2\left(1+1+1+1\right)\).
So, using part (ii),
\(S=(1-2c)+(4+c)-8=-c-3\).
(iv) If \(c=-3\), then \(S=0\).
If all four roots were real, then each term \(\left(\alpha-\frac{1}{\alpha}\right)^2\), \(\left(\beta-\frac{1}{\beta}\right)^2\), \(\left(\gamma-\frac{1}{\gamma}\right)^2\), \(\left(\delta-\frac{1}{\delta}\right)^2\) would be non-negative. Since their sum is zero, each term must be zero. Hence each root would satisfy \(x=\frac{1}{x}\), so each root would be \(\pm 1\).
But the product of the roots is \(\alpha\beta\gamma\delta=-2\), whereas a product of four numbers each equal to \(\pm 1\) can only be \(\pm 1\). This is impossible, so the roots are not all real.
