Reducible polynomials similar to Sophie Germain's identity

Question

Are there known any other simple nontrivial polynomials in two or more variables that can be factored like the one in Sophie Germain's identity? $$x^4+4y^4=(x^2-2 x y+2 y^2) (x^2+2 x y+2 y^2)$$

Besides the well known (for $n \in \mathbb {Z} ^{+}$): $$x^n-y^n=(x-y) \sum _{i=1}^n x^{i-1} y^{n-i}$$ $$x^{2 n-1}+y^{2 n-1}=(x+y) \sum _{i=1}^{2 n-1} (-1)^{i-1} x^{i-1} y^{2 n-1-i}$$ And of course besides trivial ones like $5 x^2+x y^2=x (5 x+y^2)$, where all terms share a common factor.

By simple polynomial I mean a polynomial that has few terms. The less terms the better - ideally only two terms like in the Germain's polynomial or in the known ones.

Answer 1

For example, $$ x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2z)(x+\omega^2y+\omega z), $$ for a third root of unity $\omega$.

Answer 2

More formulas similar to Sophie Germain's identity can be derived from Aurifeuillean factorizations such as

$$x^6+27y^6=(x^2+3y^2)(x^2-3xy+3y^2)(x^2+3xy+3y^2),$$ $$x^{10}-3125y^{10}=(x^2-5y^2)(x^4-5x^3 y+15x^2 y^2-25x y^3+25y^4)(x^4+5x^3 y+15x^2 y^2+25x y^3+25y^4).$$

Answer 3

The roots of $\eta^3 - 3 \eta - 1 = 0$ are $$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$ We get identity $$ \color{magenta}{ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) = (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) } $$

There is a theorem: given a ternary cubic $f, \; $ write out the Hessian matrix $H.$ The entries of $H$ are homogeneous linear in the three variables. As a result, the determinant $\Delta$ of $H$ is once again a ternary cubic. If $\Delta$ is a constant multiple of $f, \; $ then $f$ factors as the product of three homogeneous linear factors, although it might be necessary to allow complex coefficients. This all applies to Dietrich's example, where he does need complex coefficients.