We have ${{x}^{3}}-1\vdots ({{x}^{2}}+x+1)$, so the rest is $1$. Let $a^2+a+1=0$, so $a^3=1$. We use the Bezout theoreme: 1) If $n=3k$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k}}+{{a}^{3k}}+1=3\ne 0$, so it is not divisible. 2) If $n=3k+1$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k+2}}+{{a}^{3k+1}}+1={{a}^{2}}+a+1=0$, so it is divisible. 3) If $n=3k+2$, result ${{a}^{2n}}+{{a}^{n}}+1={{a}^{6k+4}}+{{a}^{3k+2}}+1=a+{{a}^{2}}+1=0$, so it is divisible.