Solve $x^{2x}+27^2=54x^x$

Question

Here is what I did i consider that $x$ is a positive integer $$27^2=x^x(54-x^x)$$ so we can see the divisors of $27$ and then deduce the value of $x$

since the sum of these two divisors must be $54$

then $x^x=27$ $x=3$

and we have the functions $f(x)=x^2x+27^2$ and $g(x)=54x^x$ have only one point of intersection

so $3$ is the only positive solution

But what if $x$ is negative ?

Answer 1

Hint : Solve for $y=x^x$, it becomes a polynomial of degree 2.

Answer 2

Substitute $u=x^x$, then we get $u^2-54u+27^2=0$. Obviously $u=27$, so $x^x=27$. $x=3$ is a solution by inspection, and it is the only one, because $x^x$ is monotonously increasing on the positive integers.

Answer 3

$y=:x^x.$

$y^2 - 2\cdot 27 y +(27)^2=$

$(y-27)^2=0;$

And now?