Answer: Answer: \(S_2=8\) and \(S_8=8S_6-12S_4-72\).

We are given that \(\alpha,\beta,\gamma,\delta\) are the roots of

\(x^4-4x^2+3x-2=0.\)

We want an equation whose roots are \(\alpha^2,\beta^2,\gamma^2,\delta^2\).

Set \(y=x^2\). Then \(x^4=y^2\), so the original equation becomes

\(y^2-4y+3x-2=0.\)

For a root, \(x=\sqrt y\), hence

\(y^2-4y+3y^{1/2}-2=0.\)

Rearrange to isolate the square root:

\(3y^{1/2}=2+4y-y^2.\)

Now square both sides:

\(9y=(2+4y-y^2)^2.\)

Expanding carefully,

\((2+4y-y^2)^2=4+16y+16y^2-4y^2-8y^3+y^4.\)

So

\(9y=y^4-8y^3+12y^2+16y+4.\)

Rearranging gives

\(y^4-8y^3+12y^2+7y+4=0.\)

Therefore \(\alpha^2,\beta^2,\gamma^2,\delta^2\) are indeed the roots of

\(y^4-8y^3+12y^2+7y+4=0.\)

From this quartic, the sum of the roots is \(8\). Hence

\(S_2=\alpha^2+\beta^2+\gamma^2+\delta^2=8.\)

Now apply the quartic satisfied by \(t=\alpha^2,\beta^2,\gamma^2,\delta^2\):

\(t^4-8t^3+12t^2+7t+4=0.\)

So

\(t^4=8t^3-12t^2-7t-4.\)

Summing this over the four values of \(t\) gives

\(\alpha^8+\beta^8+\gamma^8+\delta^8=8(\alpha^6+\beta^6+\gamma^6+\delta^6)-12(\alpha^4+\beta^4+\gamma^4+\delta^4)-7(\alpha^2+\beta^2+\gamma^2+\delta^2)-4\cdot 4.\)

Hence

\(S_8=8S_6-12S_4-7S_2-16.\)

Using \(S_2=8\),

\(S_8=8S_6-12S_4-56-16=8S_6-12S_4-72.\)

This is the required result.