Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$ 
Problem Statement:-
Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$


I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.
These are the things that I have tried to turn the given equation into a quadratic equation.
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2$$ $$\text{OR}$$
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2$$
 A: Simplifying and reducing we get
$$x^4+6x^3-9x^2-162x-243=0\,\text{and}\,x\neq 3$$
Let's look for a factorisation in the form
$$x^4+6x^3-9x^2-162x-243=(x^2+ax+b)(x^2+cx+d)$$
Developing and comparing the coefficients we get
$$\begin{align}a+c=&6\\b+d+ac=&-9\\ad+bc=&-162\\bd=&-243\end{align}$$
We obviously look first for integer coefficients. We start with $-243=-3^5$ and we test the various combinations for $b,d$.
We get the following set that works
$$\begin{align}a=&-3\\b=&-9\\c=&9\\d=&27\end{align}$$
And now we're left with two quadratics
$$\begin{align}x^2-3x-9=&0\\x^2+9x+27=&0\end{align}$$
And the roots are
$$\begin{align}x_1=&3\cdot{1+\sqrt{5}\over 2}\\x_2=&3\cdot{1-\sqrt{5}\over 2}\\x_3=&3\cdot{-3+i\sqrt{3}\over2}\\x_4=&3\cdot{-3-i\sqrt{3}\over 2}\end{align}$$
A: HINT:
Using $a^2+b^2=(a-b)^2+2ab,$
$$x^2+\left(\dfrac{3x}{x+3}\right)^2=\left(x-\dfrac{3x}{x+3}\right)^2+2\cdot x\cdot\dfrac{3x}{x+3}=\left(\dfrac{x^2}{x+3}\right)^2+6\cdot\dfrac{x^2}{x+3}$$
Generalization :
For $a^2+b^2=k$
If $\dfrac{ab}{a+b}=c$ where $c$ is a non-zero finite constant,
$$\implies k=(a+b)^2-2ab=(a+b)^2-2(a+b)c$$ $$\iff(a+b)^2-2(a+b)c-k=0$$ which isa Quadratic equation in $a+b$
Can you recognize $a,b$ here?
If  $\dfrac{ab}{a-b}=c$ use  $a^2+b^2=(a-b)^2+2ab$
