Factor $8=2^{2}\times 2$. Rewrite the square root of the product $\sqrt{2^{2}\times 2}$ as the product of square roots $\sqrt{2^{2}}\sqrt{2}$. Take the square root of $2^{2}$.
$$\frac{1}{a}\times \frac{3}{2\sqrt{2}+1}$$
Rationalize the denominator of $\frac{3}{2\sqrt{2}+1}$ by multiplying numerator and denominator by $2\sqrt{2}-1$.
Consider $\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
