Prove that the equation$x^3 +3y^3 + 9z^3 = 9xyz$ has $x=y=z=0$ as its only rational solution $x^3 +3y^3 + 9z^3 = 9xyz$ has 
$x=y=z=0$
I have seen the proof by contradiction that if it has a solution in rationals (integers specifically) with  at least 1 nonzero variable, it will have an infinite number number of integer solutions (by taking in and out factors of 3 on the original solution). I do not get why this result would lead to a contradiction. Why can't the equation have an infinite number of integer solutions.
 A: Assume by contradiction this equation has a non-tivial ratonal solution.
Since the equation is homogeneous, we can multiply by the common denominator of $x,y,z$ to get a non-trivial integer solution.
Next we can divide by $gcd(x,y,z)$ to get a non-trivial integer solution with the propert $gcd(x,y,z)=1$.
Now, show that $3|x$, write $x=3x'$, then show that $3|y$, write $y=3y'$ and finally conclude that $3|z$. But this contradicts $gcd(x,y,z)=1$.
A: As $a^3 + b^3 + c^3 =3abc\implies a+b+c=0\ or\ a=b=c.$
So $x=\sqrt[3]{3}y=\sqrt[3]{9}z$ cannot give rational roots (assume one of the them  rational other become irrational expect 0).
$x+\sqrt[3]{3}y+\sqrt[3]{9}z=0$ again cannot have rational roots expect 0 .(assume 2 rational 3rd becomes irrational)
A: My method is pretty much the same as the first solution posted with a different approach to the proof.
We can see that $3|x$. On writing $x=3x_1$ and cancelling the common factor, we get $3|y$. Write $y=3y_1$ and in the same way as before, we get $3|z$. Write $z=3z_1$ 
Now, we have circled back to $3|x_1$ so $x_1=3x_2$ or $x=3^{2}x_2$. We can do this process infinite times and the end result is that, our solution becomes $x=3^{n}x_n$ (we get similar solutions for $y$ and $z$) for arbitrarily large n. This is only possible when $x=0, y=0, z=0$.  
