In $\mathbb{R}$ find the solutions of the equation $x^4-x^3-18x^2+3x+9=0$ I couldn't solve this question and hence looked at the solution which goes as follows:
$0$ is no a root of the equation. Hence:
$$x^2-x-18+\frac{3}{x}+\frac{9}{x^2}=0. \text{ So,  } 
x^2+\frac{9}{x^2}-(x-\frac{3}{x})-18=0$$
I state that $y=x-\frac{3}{x}$, so we have that $y^2=x^2+\frac{9}{x^2}-6$, in other words $x^2+\frac{9}{x^2}=y^2+6$. Hence the equation is written as $y^2+6-y-18=0$ so $y^2-y-12=0$, hence $y=4$ or $y=-3$ so $x=2\pm\sqrt{7}$, or $x=\frac{-3\pm\sqrt{21}}{2}$
Could you please explain to me why the solution author of the solution thought of originally dividing by the equation by x and after that substituting $x-\frac{3}{x}$ with $y$? Also if you can think of a more intuitive approach could you please show it?

For the factorization, using the long, long way ,consider
$$(x^2+ax+b)(x^2+cx+d)-(x^4-x^3-18x^2+3x+9)=0$$ Expand and group terms to gat
$$(b d-9)+x (a d+b c-3)+x^2 (a c+b+d+18)+x^3 (a+c+1)=0$$ So, $c=-a-1$; replace
$$(b d-9)+x ((-a-1) b+a d-3)+x^2 ((-a-1) a+b+d+18))=0$$ So, $b=a^2+a-d-18$; replace
$$\left(d \left(a^2+a-d-18\right)-9\right)+x (-a (a (a+2)-2
   d-17)+d+15)=0$$ So, $d=\frac{a^3+2 a^2-17 a-15}{2 a+1}$; replace
$$\frac {(a-3)(a+4)(a^4+2 a^3-23 a^2-24 a-3 )} {(1+2a)^2 }=0$$ The quartic does not show any real root : so $a=3\implies c=-4$ or $a=-4\implies c=3$. Choose the one you prefer and run back for $b$ and $d$.
Edit
I made a mistake when I wrote "the quartic does not show any real root". What I was supposed to write is that "the quartic does not show any rational root". This correction followed  @Macavity's comment.
