Prove that $x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}$ 
Let $p > 3$ be a prime, and let $x,y$ be integers such that $\gcd(x,y) = 1$. Prove that $$x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}.$$

I thought about expanding $(x+y)^p$ using the binomial theorem, but I didn't see how that would help because the modulus is an algebraic expression. How can we prove the statement?

Expanding on previous comment: $\,(x+1)^p-x^p-1\,$ is divisible by $\,x^2+x+1\,$ as polynomials.
Let $\,\omega_1,\omega_2\,$ be the roots of $x^2+x+1\,$, and note that they are the two complex cube roots of unity since $x^3-1=(x-1)(x^2+x+1)$. By Vieta's relations $1+\omega_1+\omega_2=0\,$. It also follows that $1+\omega_1^p+\omega_2^p=0$ for any $p \ge 1$ which is not a multiple of $3\,$.
Then,  $\,P(\omega_1)=(\omega_1+1)^p-\omega_1^p-1=(-\omega_2)^p-\omega_1^p-1=-\omega_2^p-\omega_1^p-1=0\,$ for any odd $p \gt 3$ which is not a multiple of $3\,$. The same goes to show that $\,P(\omega_2)=0\,$, so $\,P(x)\,$ is a polynomial multiple of $\,(x-\omega_1)(x-\omega_2)=x^2+x+1\,$.
Let $\,(x+1)^p-x^p-1=Q(x)(x^2+x+1)\,$ where $\,Q(x)\,$ is a polynomial of degree at most $p-2$ with integer coefficients. Then:
$$\left(\frac{x}{y}+1\right)^p-\frac{x^p}{y^p}-1=Q\left(\frac{x}{y}\right)\left(\frac{x^2}{y^2}+\frac{x}{y}+1\right)$$
Multiplying by $\,y^p\,$ gives:
$$(x+y)^p-x^p-y^p=y^{p-2}Q\left(\frac{x}{y}\right) \cdot \left(x^2+xy+y^2\right)$$
Quite obviously, $\,y^{p-2}Q\left(\cfrac{x}{y}\right)\,$ is an integer whenever $x,y$ are integers.
As a note, the above only used the conditions that $p > 3$ must be an odd integer not divisible by $\,3\,$ (though not necessarily a prime).
