cubic equation.....

Lendy calculations giving me 186..

Have to search for a shorter method.

Thanks Rishab, Sigma

\hspace{-16}$The Equation $\bf{x^3-3x+1=0}$ has a Roots $\bf{a\;,b\;,c}$\\\\\\ Then Using Vieta,s Formula\\\\\\ $\bf{a+b+c=0}$\\\\\\ $\bf{ab+bc+ca=-3}$\\\\\\ $\bf{abc=-1}$\\\\\\ If $\bf{a+b+c=0\Leftrightarrow a^3+b^3+c^3=3abc}$\\\\\\ So $\bf{a^3+b^3+c^3=-3\Leftrightarrow \sum_{cyclic(a,b,c)}a^3=-3}$\\\\\\ Similarly $\bf{(a+b+c)^2=\sum_{cyclic(a,b,c)}a^2+2(ab+bc+ca)}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^2=6}$\\\\\\ Now Given $\bf{x^3=3x-1}$\\\\\\ So $\bf{x^5=3x^3-x^2}$\\\\\\

\hspace{-16}$So $\bf{\sum_{cyclic(a,b,c)}a^5=3.\sum_{cyclic(a,b,c)}a^3-\sum_{cyclic(a,b,c)}a^2}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^5=3(-3)-6=15}$\\\\\\ Similarly $\bf{x^3=(3x-1)\Leftrightarrow x^6=9x^2+1-6x}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^6=9.\sum_{cyclic(a,b,c)}a^2-6.\sum_{cyclic(a,b,c)}a+3=9.(6)+3=57}$\\\\\\ So $\bf{x^3=3x-1\Leftrightarrow x^8=3x^6-x^5}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^8=3.\sum_{cyclic(a,b,c)}a^6-\sum_{cyclic(a,b,c)}a^5}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^8=3.(57)-(-15)=186}$