Octal Summation

Let's manipulate the expression a little:

$x^3=x-1 \\ x^8=x^6-x^5 \\ x^8=x^3(x^3-x^2) \\ x^8=(x-1)(x-1-x^2) \\ x^8=2x^2-2x-x^3-1 \\ x^8=2x^2-2x-(x-1)-1 \\ x^8=2x^2-3x+2$

Now, by Vieta's formulas we know that $A+B+C=0$ and $AB+AC+BC=-1$ , so

$A^8+B^8+C^8=2(A^2+B^2+C^2)-3(A+B+C)+2(3) \\ A^8+B^8+C^8=2((A+B+C)^2-2(AB+AC+BC))-3(A+B+C)+6 \\ A^8+B^8+C^8=2((0)^2-2(-1))+6 \\ A^8+B^8+C^8=\boxed{10}$

By Vieta's formula , the Newton's sums or identities are $S_1 = A+B+C = 0$ , $S_2 = AB+BC+CA = -1$ , and $S_3=ABC = -1$ . Let $P_n = A^n+B^n+C^n$ , where $n$ is a natural number. We need to find $P_8$ . Then we have:

$\begin{aligned} P_1 & = S_1 = 0 \\ P_2 & = S_1P_1 - 2S_2 = 0 -2(-1) = 2 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 0 -(-1)(0) +3(-1) = -3 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 0 + 2-0 = 2 \\ P_5 & = P_3 - P_2 = -3-2=-5 \\ P_6 & = P_4 - P_3 = 2+3 =5 \\ \implies P_8 & = P_6-P_5 = 5+5 = \boxed{10} \end{aligned}$

$x^3-x+1=0 \\ x^3=x-1 \\ x^9=x^3-3x^2+3x-1 \\ x^8=x^2-3x+3-\dfrac{1}{x}.....(1) \\ ~~\\$
Now, by Vieta's formulas we know that A+B+C=0 and AB+AC+BC=−1.
$\therefore A^2+B^2+C^2= 0^2-(-2)=2........\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}= 1.\\From ~~(1)\therefore A^8+B^8+C^8\\=A^2-3A+3-\dfrac{1}{A}+B^2-3B+3-\dfrac{1}{B}+C^2-3C+3-\dfrac{1}{C}\\\large=A^2+B^2+C^2-3(A+B+C)+3*3-\left \{ \dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right \}\\= 2+0+9-1= \large \boxed{\color{#3D99F6}{ 10 }}$