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 $2^{2}$.
$$\frac{2\sqrt{2}}{3+2\sqrt{8}}$$
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 $2^{2}$.
$$\frac{2\sqrt{2}}{3+2\times 2\sqrt{2}}$$
Multiply $2$ and $2$ to get $4$.
$$\frac{2\sqrt{2}}{3+4\sqrt{2}}$$
Rationalize the denominator of $\frac{2\sqrt{2}}{3+4\sqrt{2}}$ by multiplying numerator and denominator by $3-4\sqrt{2}$.
Consider $\left(3+4\sqrt{2}\right)\left(3-4\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
