How to find $α^2(β^4 +γ^4 +δ^4)+β^2(γ^4 +δ^4 +α^4)+γ^2(δ^4 +α^4 +β^4)+δ^2(α^4 +β^4 +γ^4)$

Question

How to do the part (iv) . Please help.

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Here are my answers to the first parts:

(i)

α a root of given equation $\implies \alpha^4-5 \alpha^2 + 2 \alpha -1 = 0$

$\implies \alpha^{n+4} - 5 \alpha^{n+2} + 2\alpha^{n+1} -\alpha^n=0$

Summing over $α, \beta , \gamma , \sigma$, leads to

$S_{n+4}– 5S_{n + 2} + 2S_{n +1}- S_n=0$

(ii)

$S_2=10$

$S_4 = 5S_2 – 2S_1 + 4 = 50 – 0 + 4 = 54$

(iii)

$S_{-1} = 2$ from $y^4 – 2y^3+5y^2-1=0$

$S_3 = 5S_1 – 2S_0 + S_{–1} = –6$

$S_6 = 5S_4 – 2S_3 + S_2 = 292$

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I have no clue how to (iv)

Answer 1

Note that $\beta^4 + \gamma^4 + \delta^4 = S_4 - \alpha^4$, and similarly for the other quantities in parentheses. Substitute this in your desired expression, and expand it. You can then write the entire expression in terms of various $S$'s.