Answer: The value of \(c\) is \(2\).
A cubic equation with roots \(\alpha+\beta,\ \beta+\gamma,\ \gamma+\alpha\) is \(y^{3}-4y^{2}+y+2=0\).
A cubic equation with roots \(\dfrac{1}{\alpha+\beta},\ \dfrac{1}{\beta+\gamma},\ \dfrac{1}{\gamma+\alpha}\) is \(2z^{3}+z^{2}-4z+1=0\).
Hence \(\dfrac{1}{(\alpha+\beta)^{2}}+\dfrac{1}{(\beta+\gamma)^{2}}+\dfrac{1}{(\gamma+\alpha)^{2}}=\dfrac{17}{4}.\)
For \(x^{3}-2x^{2}-3x+4=0\) with roots \(\alpha,\beta,\gamma\), Vieta's formula gives
\(\alpha+\beta+\gamma=2\).
So \(c=2\).
Using the substitution \(y=c-x\), we have \(y=2-x\), so \(x=2-y\).
Substitute into the original equation:
\((2-y)^{3}-2(2-y)^{2}-3(2-y)+4=0\).
Expand and simplify:
\(8-12y+6y^{2}-y^{3}-2(4-4y+y^{2})-6+3y+4=0\)
\(8-12y+6y^{2}-y^{3}-8+8y-2y^{2}-6+3y+4=0\)
\(-y^{3}+4y^{2}-y-2=0\).
Multiplying by \(-1\),
\(y^{3}-4y^{2}+y+2=0\).
Since \(y=2-x=(\alpha+\beta+\gamma)-x\), when \(x=\alpha,\beta,\gamma\), the corresponding values of \(y\) are \(\beta+\gamma,\gamma+\alpha,\alpha+\beta\). So this is the required cubic.
Now let \(z=\dfrac{1}{y}\). Then \(y=\dfrac{1}{z}\), and substituting into
\(y^{3}-4y^{2}+y+2=0\)
gives
\(\dfrac{1}{z^{3}}-\dfrac{4}{z^{2}}+\dfrac{1}{z}+2=0\).
Multiplying through by \(z^{3}\):
\(1-4z+z^{2}+2z^{3}=0\),
so
\(2z^{3}+z^{2}-4z+1=0\).
This has roots \(\dfrac{1}{\alpha+\beta},\dfrac{1}{\beta+\gamma},\dfrac{1}{\gamma+\alpha}\).
Let these roots be \(r_1,r_2,r_3\). From
\(2z^{3}+z^{2}-4z+1=0\),
Vieta's formula gives
\(r_1+r_2+r_3=-\dfrac{1}{2}\),
and
\(r_1r_2+r_2r_3+r_3r_1=\dfrac{-4}{2}=-2\).
We need \(r_1^2+r_2^2+r_3^2\). Using
\(r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)\),
we get
\(r_1^2+r_2^2+r_3^2=\left(-\dfrac12\right)^2-2(-2)=\dfrac14+4=\dfrac{17}{4}\).
Hence
\(\dfrac{1}{(\alpha+\beta)^{2}}+\dfrac{1}{(\beta+\gamma)^{2}}+\dfrac{1}{(\gamma+\alpha)^{2}}=\dfrac{17}{4}.\)
