Is this identity $ (1+6x^3+9x^4)^3+(1-6x^3+3x-9x^4)^3+(1-9x^3-6x^2)^3=9x^2+9x+3$ known with $x$ is an integer?

Question

It is known that, let $x$ arbitrary integer $$(9x^4)^3+(3x-9x^4)^3+(1-9x^3)^3=1\tag{1}$$ discovred by Kurt Mahler in 1936, and $$(1+6x^3)^3+(1-6x^3)^3+(-6x^2)^3=2\tag{2}$$ discovred by A. S. Verebrusov in 1908 , Now if sum $(1)$+$(2)$ we have $0 \bmod 3$ , The nice result if we sum terms of $(1)$+$(2)$ sides by sides we obtaine a quadratique polynomial which is $ 0 \bmod 3$ then: $$ (1+6x^3+9x^4)^3+(1-6x^3+3x-9x^4)^3+(1-9x^3-6x^2)^3=9x^2+9x+3\tag{3}$$ I ask now whether the last identity $(3)$ known before ? and when it is a perfect square ?

Answer 1

There are a lot of algebraic identities out there and this one may not have been studied before. And, as remarked in the comments, it is never a square for integer $x$.

However, to make your identity look more elegant, just add $3x^3$ to both sides,

$$ (1+6x^3+9x^4)^3+(1-6x^3+3x-9x^4)^3+(1-9x^3-6x^2)^3+3x^3=3(x+1)^3$$

and you get cubes again.