<s>[INST] Prove $x^3+3y^3+9z^3-9xyz=1$ has infinity integers solutions! A question from Alibaba Global Mathematics Competition (number theory)
Prove  $\displaystyle x^{3} +3y^{3} +9z^{3} -9xyz=1$ has infinitely many integer solutions.
The hint for the question is to transform the left side to a complex polynomial.
I found that:
Set $\lambda =e^{j2\pi /3}$ then the equation can be transformed to:
$$ \left( x+3^{1/3} y+3^{2/3} z\right)\left( x+3^{1/3} \lambda y+3^{2/3} \lambda ^{2} z\right)\left( x+3^{1/3} \lambda ^{2} y+3^{2/3} \lambda z\right)=1$$
but I don't know how to continue.
 [/INST] this is my anser
first suppose we have an answer x,y,z , for simplicity,
set  $\displaystyle a=x,b=3^{1/3} y\ ,c=3^{2/3} z$
and take both sides pow3 ,we get
$\displaystyle ( a+b+c)^{3}\left( a+\lambda b+\lambda ^{2} c\right)^{3}\left( b+\lambda ^{2} b+\lambda c\right)^{3}$=1
⟹
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\left( a^{3} +b^{3} +c^{3} +3a^{2} b+3ab^{2} +3a^{2} c+3ac^{2} +3b^{2} c+3bc^{2} +6abc\right) *\\
\left( a^{3} +b^{3} +c^{3} +3\lambda a^{2} b+3\lambda ^{2} ab^{2} +3\lambda ^{2} a^{2} c+3\lambda ac^{2} +3\lambda b^{2} c+3\lambda ^{2} bc^{2} +6abc\right) *\\
\left( a^{3} +b^{3} +c^{3} +3\lambda ^{2} a^{2} b+3\lambda ab^{2} +3\lambda a^{2} c+3\lambda ^{2} ac^{2} +3\lambda ^{2} b^{2} c+3\lambda bc^{2} +6abc\right) \ =1
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
set\ A=a^{3} +b^{3} +c^{3} +6abc\ ,\ B=3a^{2} b+3b^{2} c+3ac^{2} \ ,C=3ab^{2} +3a^{2} c+3bc^{2} ,\ we\ get\\
( A+B+C)\left( A+\lambda B+\lambda ^{2} C\right)\left( A+\lambda ^{2} B+\lambda C\right) =1\\
set\ A'=A\ ,\ \ B'=B/\left( 3^{1/3}\right) ,\ \ \ C’=C/\left( 3^{2/3}\right)\\
we\ get\\
\\
\ \ \left( A'+3^{1/3} B'+3^{2/3} C'\right)\left( A'+3^{1/3} \lambda B'+3^{2/3} \lambda C'\right)\left( A'+3^{1/3} \lambda ^{2} B'+3^{2/3} \lambda C'\right) =1\\
\\
then:\\
A'\ =a^{3} +b^{3} +c^{3} +6abc\ \ is\ interger\\
B'=B/\left( 3^{1/3}\right) =\left( 3a^{2} b+3b^{2} c+3ac^{2}\right) /3^{1/3} =3x^{2} y+9y^{2} z+9xz^{2} \ is\ interger\\
C'=\left( 3ab^{2} +3a^{2} c+3bc^{2}\right) /3^{2/3} =3xy^{2} +3x^{2} z+9yz^{2} \ is\ interger\\
\\
so\ we\ get\ \ A^{\prime 3} +3B^{\prime 3} +9C^{\prime 3} -9A'B'C'=1\ such\ that\ A' >a,B' >b,C' >c\\
we\ can\ repeat\ the\ process\ to\ get\ infinity\ solutions!
\end{array}$
</s>