Let $P(x)=(x+1)^{2k}+x^{2k}+1$, $Q(x)=x^2+x+1$ then roots of equation $Q(x)=0$ is $x_1=e^{\frac{2 \pi i}{3}}$ and $x_2=e^{\frac{4 \pi i}{3}}$. But $P(x_1)= (e^{\frac{\pi i}{3}})^{2k}+(e^{\frac{2 \pi i}{3}})^{2k}+1=Q(e^{\frac{2k \pi i}{3}})=0$(from $(3;k)=1$. Similarly $(P(x_2)=0$.=> $P(x)$ divides by $Q(x)$ with $0$ residue.

Let $P(x)=Q(x)R(x)$ easy to see, that coefficients of $R(x)$ is integers Click to reveal hidden text

and degree of $R(x)$ is $2k-2$=> $R(x)=\sum_{i=0}^{2k-2} a_ix^i$ where $a_i$ is integer for all $0 \leq i \leq 2k-2$ Then $P(\frac{a}{b})=Q(\frac{a}{b})Q(\frac{a}{b})$ <=> $b^{2k}P(\frac{a}{b})=b^2Q(\frac{a}{b})\bigl( b^{2k-2} R(\frac{a}{b}) \bigr)$ <=> $(a+b)^{2k}+a^{2k}+b^{2k}=(a^2+ab+b^2)(\sum_{i=0}^{2k-2} a_i*a^i*b^{2k+2-i})$, obvious, that $(\sum_{i=0}^{2k-2} a_i*a^i*b^{2k+2-i})$ is integer ($a_i, a, b$ is integers)=> $QED$