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?

Starting with what you have written we quickly see that $$0=x^3-18x^2+81x-8=(x-8)(x^2-10x+1)$$
It is easy to see that $x\neq 8$ so we must have $\boxed {x^2-10x+1=0}$ 
Now, go back to the original equation.  Multiply by $x$ to get $$x^2-\sqrt {8x}=9x\implies x-\sqrt {8x}=10x-x^2$$
But the boxed equation tells us that $10x-x^2=1$ so $$\boxed {x-\sqrt {8x}=1}$$
