How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$ Let $x,y,z$ rationals Show that if $(x+y+z)^3=9(x^2y+y^2z+z^2x)$ then $x=y=z$
I tried this :
Let $x$ be the smallest variable 
Write $y=a+x$ and $z=b+x$ Prove $a=b=0$
 by factoring the equation as the sum of three squares.
any suggestions?

First of all $x=y=z$ is a solution to the equality, so we need to prove that all solutions other than this are not valid.
Let $y=a+x$ and $z=b+x$ where $b,a \in Q$
$(x+y+z)^{3} = 9(x^{2}y + y^{2}z + z^{2}x)$
$\Rightarrow (a+b+3x)^{3} = 9(x^{2}(a+x) + (a+x)^{2}(b+x) + (b+x)^{2}x)$
By equating coefficients, we get,
$\Rightarrow (a+b)^{3} = 9(a^{2}b)$
Now let $m= \frac{b}{a}$
$\Rightarrow m^{3} + 3m^{2} - 6m + 1=0$
Let $f(m)= m^{3} + 3m^{2} - 6m + 1$
Now, here is some food for thought,
If we can prove that $b$ or $a$ is irrational , we would arrive at a contradiction because $x,y,z$ are rational numbers. 
If $\frac {b}{a}$ is irrational then either $b$ or $a$ will be irrational and we will arrive at a contradiction, which would give us no solution other than $x=y=z$.
So, all we need to do is prove that the cubic $f(m)=0$ has no rational roots.
$m^{3} + 3m^{2} - 6m + 1 = 0$
Let's assume that $f(m)=0$ has rational roots.
Substitute $m=\frac {p}{q}$ , where $gcd(p,q)=1$ and $p,q \in I$
$(\frac {p}{q})^{3} + 3(\frac {p}{q})^{2} - 6(\frac {p}{q}) + 1=0$
$\Rightarrow p(p^{2} + 3pq - 6q^{2}) = -q^{3}$
We know that $gcd(p,q)=1 \Rightarrow gcd(p,q^{3})=1$.
So $p$ must divide $-1$ 
Therefore the possible values of $p$ are $\pm 1$
Similarly we can write ,
$\Rightarrow q(q^{2} - 6pq + 3p^{2}) = -p^{3}$
So $q$ must divide $-1$ 
Therefore the possible values of $q$ are $\pm 1$
The possible roots for $f(m)=0$ are $\frac{p}{q} = \pm 1$.
But $f(1)=-1$ and $f(-1)=9$ .
Therefore, $f(m)=0$ has no rational roots.
Therefore, $\frac {b}{a}$ is irrational which is a contradiction.
Hence , $x=y=z$ is the only solution.
