Greatest common divisor of $(x+1)^{4n+3} + x^{2n}$ and $x^3-1$. I have to find the greatest common divisor of 
$$(x+1)^{4n+3} + x^{2n}$$
and
$$x^3-1$$
I know I can express the second polynomial as:
$$x^3-1 = (x-1)(x^2+x+1)$$
So I would have to check if the first polynomial is divisible by $(x^3-1)$, $(x^2+x+1)$ or $(x-1)$ and if it is not divisible by any of those, then the two polynomials do not have a common divisor except for $1$. But I don't know how I can divide the polynomial
$$(x+1)^{4n+3} + x^{2n}$$
by those $3$ other polynomials and therefore can't check the greatest common divisor.
 A: Only $x^2+x+1$ can be an answer because $x-1$ is not valid.
Modulo $x^2+x+1$ we obtain: 
$$(x+1)^{4n+3}+x^{2n}\equiv x^{2n}-x^{8n+6}=x^{2n}(1-(x^3)^{2n+2})\equiv0.$$
A: Let $Q(x) = (x+1)^{4n+3} + x^{2n}$. Then $Q(1) = 2^{4n+3} + 1 \neq 0$, hence $x-1$ does not divide $Q(x)$. 
We have $$Q(x) = (x+1)^{4n+3} + x^{2n} = (x+1)[(x+1)^2]^{2n+1} + x^{2n} =$$
$$= (x+1)[(x^2+x+1) + x]^{2n+1}+ x^{2n} = $$
$$=(x^2+x+1)P(x) + (x+1)x^{2n+1} + x^{2n} = (x^2+x+1)(P(x) + x^{2n}),$$
hence $x^2+x+1$ divides $Q(x)$.
A: Hint $\,\ x\!-\!1\nmid f(x)\,$ by $\,f(1)\neq 0,\,$ but $\ x^2\!+\!x\!+\!1\mid f(x)\,$ by
$\!\!\!\begin{align}\bmod\, \color{#0a0}{x^2\!+\!x\!+\!1}\!:\,\ f(x)\,\equiv\ &x^{\large 2n}+(\color{#0a0}{x\!+\!1})^{\large 4n+3}\\[.2em]
\equiv\  &x^{\large 2n}+({\color{#0a0}{-x^{\large 2}}})^{\large 4n+3}\ \  {\rm thus\ reducing\ using}\ \  x^{\large\color{#c00}3}\equiv 1\\[.2em]
\equiv\ &x^{\large 2n}\,  -\, x^{\large\color{#90f}{2n}}\equiv 0,\ \, {\rm by}\ \ \color{#0a0}2(4n\!+\!3)\equiv\color{#90f}{2n}\!\!\!\pmod{\!\color{#c00}3}\end{align}$
