jmc

Let $\omega$ satisfy $x^2+x+1=0,$ so ${\omega }^2+\omega +1=0.$ Then $(\omega -1)({\omega }^2+\omega +1)={\omega }^3-1=0,$ so ${\omega }^3=1.$ Also, $$\begin{array}{rl}{\omega }^{10}+{\omega }^5+1& ={\omega }^9\cdot \omega +{\omega }^3\cdot {\omega }^2+1\\ & =\omega +{\omega }^2+1\\ & =0.\end{array}$$Therefore, $x^2+x+1$ is a factor of $x^{10}+x^5+1.$

To bring out this factorization, we can write $$\begin{array}{rl}{x}^{10}+x^5+1& =x^{10}-x+x^5-x^2+x^2+x+1\\ & =x(x^9-1)+x^2(x^3-1)+x^2+x+1\\ & =x(x^3-1)(x^6+x^3+1)+x^2(x-1)(x^2+x+1)+x^2+x+1\\ & =x(x-1)(x^2+x+1)(x^6+x^3+1)+x^2(x-1)(x^2+x+1)+x^2+x+1\\ & =\boxed{(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)}.\end{array}$$