Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational Prove $x^4-18x^2+36x-27$ can never be a square rational (excluding 0), when x is rational

I have tried to use modulus, but didn't get anywhere, any help would be greatly appreciated.
The equation $y^2=x^4-18x^2+36x-27$ is birationally equivalent to the elliptic curve $w^2=z^3-432$, with $x=\frac{w-36}{2(z-12)}$. This curve has rank $0$ and torsion group $\mathbb Z/3\mathbb Z$, hence only two rational points $(z,w)=(12,\pm36)$. If we substitute these points into the formula for $x$, the positive $w$ gives the excluded $x=3$ solution and the negative $w$ incurs a division by zero. Hence $x^4-18x^2+36x-27$ can never be a nonzero rational square for rational $x$.
Say $x^4-18*x^2+36*x-27 = n^2$

The polynomial factors up

$(x-3)*(x^3+3*x^2-9*x+9) = n^2$

It means that if $x-3$ is a perfect square, then $x^3+3*x^2-9*x+9
$ is also a perfect square or both aren't perfect square

Just like $36=9×4=12×3$

If  $x-3$ is not a perfect square, then polynomial $x^3+3*x^2-9*x+9$ is not also a perfect, therefore it must factor up further, Just like $36=12×3=4×3×3$

But $x^3+3*x^2-9*x+9$ does not factor

It doesn't factor as supposed, it has to factor into $(x-3)*(x-a)^2$ in my definition, but there's no value of $a$ to fix this therefore $x^3+3*x^2-9*x+9$ does not factor to $(x-3)*(x-a)^2$

Say $x-3 = k^2$, $x = 3+k^2$

$(3+k^2)^3+3*(3+k^2)^2-9*(3+k^2)+9 = (n/k)^2$

$k^6+12*k^4+36*k^2+36 = (n/k)^2$

$k^2*(k^2+6)^2+36 = (n/k)^2$

$(k\cdot(k^2+6))^2 + 6^2 = (n/k)^2$

It's easy to see that no integer $k$ exist to that make this a perfect square

Therefore $x^4-18*x^2+36*x-27$ is not a perfect square