Can we generalize this formula We have these identities :
$$x^{2n+1} + y^{2n+1} = (x+y)(x^{2n} - x^{2n-1}y + \cdots -y^{2n-1}x + y^{2n})$$
and
$$x^{2n+1} - y^{2n+1} = (x-y)(x^{2n} + x^{2n-1}y + \cdots + y^{2n-1}x + y^{2n})$$
If $n=1$ or $n=2$ , It is obvious that second parenthesis is non-negative .
For example for $n=1$ :
$x^2 - xy + y^2 \ge 0 \iff 2x^2 -2xy + 2y^2 \ge 0 \iff (x-y)^2 +x^2 + y^2 \ge 0 \iff \begin{cases} (x-y)^2 \ge 0 &  \\ x^2\ge0 \\ y^2 \ge 0  \end{cases} $
Can we prove that for all $n$ in these identities second parenthesis is non-negative ?

The left side and the first term on the right will always have the same sign, so the second term on the right can't be negative.
Edit in response to comment asking for more details.
Since the function $x \mapsto x^{\text{odd}}$ is an increasing function the left side of the second equation has the same sign as the first term on the right.
For the first equation the assertion is obvious if $x$ and $y$ have the same sign. If they have opposite signs then the signs are the same as the sign of whichever of $x$ and $y$ is further from $0$.
