<s>[INST] Roots of An Equation - Finding Equations in other degrees of roots (Link: https://pastpapers.papacambridge.com/viewer/caie/cambridge-advanced-as-and-a-level-mathematics-9231-2021-may-june-9231s21gtpdf-9231s21ms11pdf-9231s21ms12pdf-9231s21ms13pdf-9231s21ms21pdf-9231s21ms22pdf-9231s21ms23pdf-9231s21ms31pdf-9231s21ms32pdf-9231s21ms33pdf-9231s21ms41pdf-9231s21ms42pdf-9231s21ms43pdf-9231s21qp11pdf)
This is from a paper, so I'll give a summary of the question first.
$x^{4}-2x^{3}-1=0$
has roots $\alpha ,\beta ,\gamma , \delta$
a) Find equation with $\alpha^{3} ,\beta^{3} ,\gamma^{3} , \delta^{3}$
Answer $y = x^{3}$
Therefore, $y^{4}-8y^{3}-12y^{2} - 6y - 1 = 0$
b) Find the value of  $1/\alpha ^{3}+ 1/\beta ^{3} + 1/\gamma ^{3} + 1/\delta ^{3}$
Answer -6
*** c) Find the value of $\alpha ^{4}+ \beta ^{4} + \gamma ^{4} + \delta ^{4}$ ***
Now the mark scheme says
$\alpha ^{4}+ \beta ^{4} + \gamma ^{4} + \delta ^{4} = 2(\alpha ^{3}+\beta ^{3}+\gamma^{3}+\delta ^{3}) + 4$ which is = 20. The mark scheme also says Uses original equation
Can someone explain to me how they got this equation for question C? Thanks in advance.
 [/INST] (From dxiv)
Original Equation $x^4 - 2x^3 - 1 = 0$
Therefore $x^4 = 2x^3 +1 $
Using each of the roots instead of x, we get
$\alpha^4 = 2\alpha^3 +1$
.....
.....
$\delta^4 = 2\delta^3 +1$
Add these all together to get $\alpha^4+\beta^4+\gamma^4+\delta^4 = 2(\alpha^3+\beta^3+\gamma^3+\delta^3) + 4$
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