Substitution is blind - 3

Given that $x = 9 + 4\sqrt 5$ , $\implies \dfrac 1x = \dfrac 1{9+4\sqrt 5} = \dfrac {9-4\sqrt 5}{(9-4\sqrt 5)(9+4\sqrt 5)} = 9 - 4\sqrt 5$ .

Then we have:

$\begin{aligned} x^2 - \frac 1{x^2} & = \left(x+\frac 1x\right) \left(x-\frac 1x\right) \\ & = \left(9+4\sqrt 5 + 9 - 4\sqrt 5 \right)\left(9+4\sqrt 5 - 9 + 4\sqrt 5 \right) \\ & = 18 \times 8\sqrt 5 \\ & \approx \boxed{322} \end{aligned}$

$x=9+4\sqrt{5} = \sqrt{81}+\sqrt{80}$ $\implies \dfrac{1}{x} = \sqrt{81} -\sqrt{80}$

$x^2 -\dfrac{1}{x^2} = (\sqrt{81}+\sqrt{80})^2 + ( \sqrt{81} -\sqrt{80})^2$

Let $a= \sqrt{81}$ and $b= \sqrt{80}$

$\implies x^2 -\dfrac{1}{x^2} = (a+b)^2 - (a-b)^2 = a^2 +b^2 +2ab -a^2 -b^2 +2ab =4ab =4 \times \sqrt{81} \times \sqrt{80} = 4 \times 9 \times 4\sqrt{5} = 144 \sqrt{5} \approx \textcolor{#20A900}{\boxed{322}}$