Quartic polynomial roots

Question

The equation $x^4 - x^3-1=0$ has roots α, β, γ, δ. By using the substitution $y=x^3$ find the exact value of $α^6+β^6+γ^6+δ^6$ .

The solution is

$x=y$ ^(1/3)

$y^4=(1+y)^3$

$y^4 -y^3 -3y^2-3y^2-1=0$

$S$_N+4 $=$ $S$_N + $S$_N+3

$S$_-1 = $\frac {0}{1} =0$

$S$₂ = $ 1^2 -2*0 =1$

$S$₃ = $0+1 =1$

$S$₄ = $1+4 =5$

$S$₅ = $5+1=6$

$S$₆ = $6+1 =7$

Therefore, $α^6+β^6+γ^6+δ^6 = 7$

Can someone explain what is this $S$_n is and how did they work out solution from that.

Answer 1

Multiply each term by $x^n$

$\pmb S_n =x^n$

Then it becomes

$S$_N+4 $=$ $S$_N + $S$_N+3

Answer 2

This looks like an application of the Newton identities which are a generalization of Viete's identities connecting roots and coefficients. If $S_k=\sum_{j=1}^d α_j^k$ is the degree $k$ power sum of the roots of the polynomial equation $0=x^4+c_1x^3+c_2x^2+c_3x+c_4$, then $$ c_1+S_1=0\\ 2c_2+c_1S_1+S_2=0\\ 3c_3+c_2S_1+c_1S_2+S_3=0\\ 4c_4+c_3S_1+c_2S_2+c_1S_3+S_4=0\\ c_4S_1+c_3S_2+c_2S_3+c_1S_4+S_5=0\\ c_4S_2+c_3S_3+c_2S_4+c_1S_5+S_6=0\\ $$

For the given polynomial with $(c_1,c_2,c_3,c_4)=(-1,0,0,-1)$ this gives $$ S_1=1,\\ S_2=S_1=1,\\ S_3=S_2=1,\\ S_4=4+S_4=5,\\ S_5=S_1+S_5=6,\\ S_6=S_2+S_6=7. $$

Or using the proposed approach, $S_6$ is the square sum of the roots of the polynomial for $y$ which was found as $$0=y^4-y^3-3y^2-3y-1=y^4+C_1y^3+C_2y^2+C_3y+C_4,$$ where again one finds $S_3=-C_1=1$ and $$S_6=-C_1S_3-2C_2=7.$$