Polynomial divisibility exercise How do I prove that $(X+1)^{6n+1}-X^{6n+1}-1$ is divisible by $(X^2+X+1)^2$ without derivative functions? (I am 10th grade).

Let us denote $q_n(x)=(x+1)^{6n+1}-x^{6n+1}-1$. We may notice that
$$ q_{n+1}(x)-(x+1)^6 q_n(x) = \left[(x+1)^6-x^6\right]x^{6n+1}+\left[(x+1)^6-1\right] $$
where both $(x+1)^6-x^6$ and $(x+1)^6-1$ are multiples of $x^2+x+1$:
$$ q_{n+1}(x)-(x+1)^6 q_n(x) = (x^2+x+1)\left[(1 + 2 x)(1 + 3 x + 3 x^2)x^{6n+1}+x (2 + x) (3 + 3 x + x^2)\right] $$
In order to prove the claim by induction, it follows that it is enough to show
$$ (x^2+x+1)\mid \left[(1+2x)x^{6n+1}+x^3(2+x)\right] $$
for any $n\geq 0$. This can be done by induction, again. Let us denote $r_n(x)=(1+2x)x^{6n+1}+x^3(2+x)$. The last claim holds for $n=0$ and 
$$ r_{n+1}(x)-x^6 r_n(x) = x^3 \left(2+x-2 x^6-x^7\right) = x^3(2+x)(1-x^6) $$
so we are done, since $x^2+x+1$ is a divisor of $1-x^6$.
