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).
 A: 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 $1-x^6$.
A: One way is to use the binomial theorem.
Assume $a$ is a root of $x^2+x+1=0$. In particular $a^3=1$ and $a+1=-a^2$.
We want to prove that we can divide two times by $y=x-a$ the polynomial $$p(x)=(x+1)^{6n+1}-x^{6n+1}-1$$ 
Replacing $x=y+a$ we get 
$$(y+(a+1))^{6n+1}-(y+a)^{6n+1}-1$$
Using the binomial theorem this is
$$\sum_{k=0}^{6n+1}\binom{6n+1}{k}y^k(a+1)^{6n+1-k}-\sum_{k=0}^{6n+1}\binom{6n+1}{k}y^ka^{6n+1-k}-1$$
The constant term is $$(a+1)^{6n+1}-a^{6n+1}-1=(-a^2)^{6n}(-a^2)-a^{6n}\cdot a-1=-a^2-a-1=0$$
The coefficient of the term of degree $1$ is 
$$(6n+1)(a+1)^{6n}-(6n+1)a^{6n}=(6n+1)[(-a^2)^{6n}-a^{6n}]=(6n+1)[1-1]=0$$
Therefore, $y^2=(x-a)^2$ divides $p(x)$. Since the computation didn't depend on which root of $x^2+x+1$ we used, it follows that $(x^2+x+1)^2$ divides $p(x)$.

In the realm of polynomials, everything that can be proved using derivatives can be proved using the binomial theorem instead.
