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?

CONCLUSION:The roots of $\eta^3 - 3 \eta - 1 = 0$ are 
$$  A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B =   2 \cos  \left( \frac{5 \pi}{9} \right)  \approx -0.347 \; \; \; , C = 2 \cos      \left( \frac{ \pi}{9} \right)  \approx 1.879 \; \; \; . $$
Note that we do have
$$ A+B+C=0 \; . \;  $$
We get identity
$$ \color{magenta}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)}  $$ 
which confirms that the surface is three planes sharing the line $x=y=z\; . \;$ 
ORIGINAL:
Take $$ x = r-s-t \; , \; \; \; y = r+s-t \; , \; \; \; z = r + 2  t \; ,  $$ so
$$ 3r = x+y+z \; \; , \; \;  2s = y - x \; , \; \; \; 6t = 2 z - x - y \; \; .  $$
Note $(x,y,z)$ is a rational triple if and only if $(r,s,t)$ is a rational triple.
Then
$$ x^2 y + y^2 z + z^2 x = 3 r^3 + \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right)  $$
and
$$ 9 \left( x^2 y + y^2 z + z^2 x\right) - (3r)^3 = 9 \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right) $$
If, for example, $t \neq 0,$ divide by $t^3$ and we must have a root of $p^3 + 3 p^2 - 9p-3$ which is irreducible. If $s \neq 0$ use the reciprocal. Insisting on rational values, we find that both $s,t$ are zero, so
$$ y-x = 0 \; , \; \; \;  2z - x - y = 0 \; , \; $$
and $$ x=y=z $$
Meanwhile, the method answers a simple question, what kind of surface are we describing in $\mathbb R^3 \; ?$ If we have an irrational root $p$ of $p^3 + 3 p^2 - 9p-3=0$ we have some other irrational real $q$ such that 
$$ 2z-x-y = q(y-x) \; , $$
$$ (q-1)x + (-q-1) y + 2 z = 0 \; \; ,  $$
which is evidently a plane containing the line $x=y=z.$ I mostly think the surface is three planes, arranged around the line $x=y=z$ at equal angles, like radii of a circle. Indeed, compared with axes given by vectors $v_1 = (-1,1,0)/ \sqrt(2)$ and $v_2 = (-1,-1,2)/ \sqrt(6),$ it seems the three planes are rotated from $v_1$ in the direction of $v_2$ at exactly the three angles $40^\circ, 100^\circ, 160^\circ,$ repetitions at  $220^\circ, 280^\circ, 340^\circ,$ so we see every $60^\circ \; .$  As $360/9 = 40$ this has a bit of plausibility. 
Next day: confirming the nature of the surface: first, it is defined by the "curve" gotten by intersecting the surface with the plane $x+y+z = 0,$ as it is a "cylinder" over that curve, with axis of translation the expected line $x=y=z.$ If $$  x = X + t \; , \; \;  y = Y + t \; , \; \; z = Z + t \; \; , $$
we find
$$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) =   (X+Y+Z)^3 - 9 \left( X^2 Y + Y^2 Z + Z^2 X \right)  $$
Wednesday, finally got it. also Tottenham just scored on Juventus in the Champions League. 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
$$ \color{red}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)}  $$ 
which confirms that the surface is three planes sharing the line $x=y=z,$ as $A+B+C = 0$
