Quartic polynomial roots 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.

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 $$S_6=-C_1S_3-2C_2=7.$$
