$d\in\Bbb Z^{\ge2}$.
What is the minimal degree of $f\in\mathbb{Q}[x,y,z]$ such that $f(x^{d-1},y^{d-1},z^{d-1})$ is divisible by $x^d+y^d+z^d$?
Let $\zeta=\exp(\frac{2\pi i}{d-1})$.
Since $f(x^{d-1},y^{d-1},z^{d-1})$ is divisible by $x^d+y^d+z^d$, by replacing $y$ with $\zeta^i$ and replacing $z$ with $\zeta^j$, we get $f(x^{d-1},(\zeta^iy)^{d-1},(\zeta^jz)^{d-1})=f(x^{d-1},y^{d-1},z^{d-1})$ is divisible by$$x^d+(\zeta^iy)^d+(\zeta^jz)^d=x^d+\zeta^{di}y^d+\zeta^{dj}z^d$$ and since $\zeta^{di}=\zeta^i,\zeta^{dj}=\zeta^j$, we get $f(x^{d-1},y^{d-1},z^{d-1})$ is divisible by $x^d+\zeta^iy^d+\zeta^jz^d$ for all $1\le i,j\le d-1$.
$x^d+\zeta^iy^d+\zeta^jz^d$ are all irreducible so $f(x^{d-1},y^{d-1},z^{d-1})$ is divisible by their product $$ \prod_{\substack{1\le i\le d-1\\1\le j\le d-1}}(x^d+\zeta^iy^d+\zeta^jz^d) $$
so $f(x,y,z)$ is divisible by the following polynomial of degree $d^2-d$. \begin{align} &\prod_{\substack{1\le i\le d-1\\1\le j\le d-1}}(x^{d/(d-1)}+\zeta^iy^{d/(d-1)}+\zeta^jz^{d/(d-1)})\\ &=\prod_{1\le i\le d-1}\left((x^{d/(d-1)}+\zeta^iy^{d/(d-1)})^{d-1}+z^d\right) \end{align} Then I only need to prove it is in $\mathbb{Q}(x,y,z)$ and irreducible.
Example:
| $d$ | $f$ of minimal degree | degree |
|---|---|---|
| $2$ | $x^2+y^2+z^2$ | $2$ |
| $3$ | $x^6+ y^6+ z^6 - 2 x^3 y^3 - 2 x^3 z^3 - 2 y^3 z^3 $ | $6$ |
| $4$ | $(x^4+y^4+z^4)^3-27x^4y^4z^4$ | $12$ |