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.

Multiply each term by $x^n$
$\pmb S_n =x^n$
Then it becomes
$S$N+4 $=$ $S$N + $S$N+3
