Consider the polynomial : $\mathcal{f(x)}=x^{10}+x^5+1$ and $\mathcal{g(x)}=x^2+x+1$. It's known that the third root of unity $\omega$ satisfies the equation, so, we can say $\mathcal{f(x)}=\mathcal{g(x)\cdot h(x)}$.
Please explain why/how the fact that the third root of unity satisfies the equation implies $\mathcal{g(x) \mid f(x)}$
Hint $\bmod\, x^{\large 2}\!+\!x\!+\!1\!:\,\ x^{\large 3}\equiv 1\,\Rightarrow\, x^{\large \color{#c00}{10}}+x^{\large\color{#0a0}5}+1\equiv x^{\large \color{#c00}1}+x^{\large \color{#0a0}2}+1\equiv 0$
Remark $ $ Generally this shows $\ x^{2}\!+\!x\!+\!1\mid x^\color{#c00}A\! +\! x^\color{#0a0}B\! +\! x^C\ $ if $ \ \{\color{#c00}A,\color{#0a0}B,C\}\equiv \{\color{#0a0}2,\color{#c00}1,0\}\pmod{\!3}.$
See this answer for further discussion and generalizations.
Note: above $x$ is an "indeterminate" not a specific root, i.e. we do polynomial arithmetic in $\,\Bbb Z[x].\,$ Comparing to Ennar's answer, instead of doing Euclidean division of $\,f\,$ by $\,x^2+x+1\,$ we do a division by its multiple $\,x^3−1\,$ which is simpler, since we need only the remainder, hence we need only compute $\,f\bmod\, x^3−1\,$ doable simply by applying the rewrite rule $\,x^3\to 1,\,$ i.e. $$\ f\bmod x^2\!+\!x\!+\!1\ =\ \underbrace{(f\bmod x^3\!-1)}_{\Large x^3\to\, 1}\bmod x^2\!+\!x\!+\!1$$
a "mod" optimization we all know subconsciously since grade school, where we frequently use
$$n\bmod 2\ = \ (n\bmod 10)\bmod 2$$
to compute the parity (even/odd) of $\,n\,$ via the parity of its units digit $= n\bmod 10$
As I explain in the linked answer, the same modular optimization works for any factor of a "simple" polynomial (e.g. a binomial or fewnomial, or one that is easy to compute modulo it)
Write $f(x) = k(x)g(x)+r(x)$ where $r$ is linear. Then if $a,b$ are a roots of $g(x)$ we have $a^3=b^3=1$ and $a,b\ne 1$. Then
$$f(a)= a^{10}+a^5+1 = a+a^2+1 =0$$ so $r(a)=0$. The same is true for $b$, so $r(b)=0$. Thus $r$ has two different zeroes and so $r\equiv 0$ and thus $g\mid f$.
