If we let S(n)=a^n+b^n+c^n, where a,b,c are the roots of the equation, then it satisfies S(n+3)=2S(n+2)+S(n+1)-S(n) S(0)=a^0+b^0+c^0=3 S(1)=a+b+c=2 S(2)=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=4^2-2(-1)=18 Therefore, S(3)=2(18)+2-3=35 Alternative method: a+b+c=2 a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=4^2-2(-1)=18 ab+bc+ca=-1 abc=-1 a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca) =2(18+1) =38 a^3+b^3+c^3-3(-1)=38 a^3+b^3+c^3=38-3=35

[Edited by Andy for better readibility. Smiles are disabled so that 8) can be displayed properly.]