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.

Hint $\,\ x\!-\!1\nmid f(x)\,$ by $\,f(1)\neq 0,\,$ but $\ x^2\!+\!x\!+\!1\mid f(x)\,$ by
$\!\!\!\begin{align}\bmod\, \color{#0a0}{x^2\!+\!x\!+\!1}\!:\,\ f(x)\,\equiv\ &x^{\large 2n}+(\color{#0a0}{x\!+\!1})^{\large 4n+3}\\[.2em]
\equiv\  &x^{\large 2n}+({\color{#0a0}{-x^{\large 2}}})^{\large 4n+3}\ \  {\rm thus\ reducing\ using}\ \  x^{\large\color{#c00}3}\equiv 1\\[.2em]
\equiv\ &x^{\large 2n}\,  -\, x^{\large\color{#90f}{2n}}\equiv 0,\ \, {\rm by}\ \ \color{#0a0}2(4n\!+\!3)\equiv\color{#90f}{2n}\!\!\!\pmod{\!\color{#c00}3}\end{align}$
