Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to? 
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?

My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?

Let $x=t^2,$ where $t>0$.
Thus, $$t^3-9t-\sqrt8=0$$ or
$$t^3+\sqrt8t^2-\sqrt8t^2-8t-t-\sqrt8=0$$ or
$$(t+\sqrt8)(t^2-\sqrt8t-1)=0$$ or
$$x-\sqrt{8x}=1.$$
