divisible large degree polynomial Let $n$ be an even positive integer and $a$, $b$ real numbers such that $b^n=3a+1$. Prove that if $(X^2+X+1)^n-X^n-a$ is divisible by $X^3+X^2+X+b$, then $a=0$ and $b=1$. 
I am thinking of using the fact that $X^2+X+1=-b/X$, but I don't see it really helping
 A: First 
If $b = 1$, then $q(x) = x^3 + x^2 + x + 1 = (x + 1)(x + i)(x − i)$. Also, $a = 0$, so $p(x) = (x^2 +x+1)^n−x^n$, and it is easy to verify that $−1$ and $\mp i$ are indeed roots of $p(x)$. Thus $q(x)|p(x)$. 
Second 
The other direction, assume that $q(x) | p(x)$. Then $q(x)$ divides $$x^np(x) = (x^3+x^2+x)^n−x^{2n}−ax^n\tag1$$ But also $q(x)$ divides 
$$(x^3+x^2+x)^n−(−b)^n = (x^3+x^2+x+b)\sum_{i=0}^{n-1}(x^3 +x^2 +x)^i(−b)^{n−1−i}\tag2$$ 
A subtraction operation between $(1)$ and $(2)$ noticing that $n$ is even, so that $(−b)^n = b^n$  gives us
$$ q(x)|(x^{2n} +ax^n −b^n)=f(x) \tag3 $$
We now consider the roots of $f(x)$ Substituting $y = x^n$, we have $f(x) = y^2 +ay−b^n=g(y)$
 Because $n$ is even, $b^n$ is nonnegative, and it follows that the discriminant of $g(y)$ is nonnegative. Thus, $g(y)$ has two roots $u$ and $v$ in $\mathbb{R}$, with $$u+v =−a, uv =−b^n \tag{4}$$ 
Finally every root of $f(x)$ verify $$x^n=u \text{ or } x^n=v\tag{5}$$
Let $\alpha,\beta,\gamma$ be the roots of $p(x)$ then :
$$\alpha\beta\gamma=-b \tag{6}$$
and we know that $q(x)|f(x)$ then $\alpha,\beta,\gamma$ verifies $(5)$, now we have $4$ cases:


*

*Two roots verify $x^n=u$ and one root verify $x^n=v$, we can assume WLOG that:
$$\alpha^n=\beta^n=u \text{ and } \gamma^n=v \tag{7}$$
now using $(6)$ we have $|\alpha||\beta||\gamma|=|b|$ and using $(4)$ and $(7)$ we have $|\alpha||\gamma|=|\beta||\gamma|=|b|$ hence : $|\alpha|=|\beta|=1$ and $|\gamma|=b$, and because $u,v$ are reals then using $(7)$ we have $u=1$ and $v=-b^n$ or $u=-1$ or $v=b^n$ (see $(4)$), in this two cases we have $u+v=\mp(1-b^n)=\mp(1-(3a+1))=\mp 3a$ and returning to $4$ again we have $a=0$ and hence $b=\mp1$, if $b=-1$ then $q(x)=x^3+x^2+x-1$ and $f(x)=x^{2n}-1$ it's easy to verify that $f(x)$ does not divide $f(x)$ because $f$ has only the two real roots $\mp1$ and $q$ has another different real root. Here we conclude that $a=0$ and $b=1$

*Two roots verify $x^n=v$ and one root verify $x^n=u$ (the same approach as in the case $(1)$

*The three roots verify $x^n=u$ , then because of $(6)$ each root has absolute value $|b|^{1/3}$, On the other hand, at least one root is real because $q(x)$ has odd degree. Thus, either $c=b^{1/3}$ or $-c=−b^{1/3}$ must be a root of $q(x)$ and hence in the first case $q(c)=2c^3 +c^2 +c = 0$ hence $c=0$(impossible) or $0<2c^2 +c+1 = 0$ (impossible),so $q(-c)=0$ and $c^2-c=0$ and because $c\neq 0$ we have $c=1$ thus $b=1$.

*The three roots verify $x^n=v$ (the same approach as in $(3)$)
