Question

Solve:

$y^3-9x^2+27x-27=0z^3-9y^2+27y-27=0x^3-9z^2+27z-27=0$

Answer 1

Adding the equation, we get
${(x-3)}^3+{(y-3)}^3{+(z-3)}^3=0$
The triplet $(3,3,3)$ is a solution of $\left(1\right)$.
It can be verified that it is also a solution of the given system of equations. (Note that verification of this statement is obligatory)
Let us now show that the system has no other solutions. From the first equation of the system, we have $y^3=9x^2-27x+27$.
The discriminant of the quadratic $9x^2-27x+27$ is negative.
Hence if $(x_0,y_0,z_0)$ is a solution of the system, then $y_0^3>0$. Consequently $y_0>0$. Similarly $z_0>0$ and $x_0>0$.
Now from the first equation of the system, we have
$(y_0-3)(y_0^2+3y_0+9)=9x_0(x_0-3).$
Since $x_0>0$ and $y^2+3y_0+9>0$.
It follows from $\left(2\right)$ that the numbers $y_0-3$ and $x_0-3$ cannot differ in sign. Similarly it follows from the second equation that $z_0-3$ and $y_0-3$ cannot have different signs either.
Thus if $x_0>3$, then $y_0>3$ and $z_0>3$; and if $x_0<3,$ then $y_0<3$, and $z_0<3$. Thus neither $x_0>3,y_0>3,,z_0>3$ nor $x_0<3,y_0<3,z_0<3$ are suitable. Hence $x_0=3,y_0=3$ and $z_0=3$ is the only solution.