Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into four possible radicals

Question

Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into possible radicals. This is an answer to an obscure closed question, here on the site. While there is an answer posted, it isn't the complete solution.

Answer 1

$$\begin{align} \sqrt{20+ \sqrt{96}+\sqrt{12}} &= \sqrt{5\cdot4+ 4\sqrt{6}+2\sqrt{3}} \\ &= \sqrt{2}\sqrt{5\cdot2+ 2\sqrt{6}+\sqrt{3}} \\ &= \sqrt{2}\sqrt{5\cdot2+\sqrt{3}\left(2\sqrt{2}+1\right)} \\ \end{align}$$

By Using the identity of denesting $\sqrt{a\pm b\sqrt{c}}=\sqrt{(a+d)/2} \pm\sqrt{(a-d)/2}$, where $d=\sqrt{a^2-b^2c}$.

$$\begin{align} \sqrt{2}\sqrt{5\cdot2+\sqrt{3}\left(2\sqrt{2}+1\right)} &= \sqrt{2}\left(\sqrt{\frac{10+\sqrt{10^{2}-3\left(2\sqrt{2}+1\right)^{2}}}{2}}+\sqrt{\frac{10-\sqrt{10^{2}-3\left(2\sqrt{2}+1\right)^{2}}}{2}}\right) \\ &= \sqrt{2}\left(\sqrt{\frac{10+\sqrt{73-12\sqrt{2}}}{2}}+\sqrt{\frac{10-\sqrt{73-12\sqrt{2}}}{2}}\right) \\ &= \sqrt{10+\sqrt{73-12\sqrt{2}}}+\sqrt{10-\sqrt{73-12\sqrt{2}}} \\ \end{align}$$

Denesting $\sqrt{73-12\sqrt{2}}$ using the above mentioned way,

$$\begin{align} \sqrt{10+\sqrt{73-12\sqrt{2}}}+\sqrt{10-\sqrt{73-12\sqrt{2}}} &= \sqrt{10+\sqrt{72}-1}+\sqrt{10-\left(\sqrt{72}-1\right)}\\ &= \sqrt{9+6\sqrt{2}}+\sqrt{11-6\sqrt{2}} \\ &= \sqrt{6}+\sqrt{3}+3-\sqrt{2} \\ \end{align}$$

Hence, $$\sqrt{20+ \sqrt{96}+\sqrt{12}}=\sqrt{6}+\sqrt{3}+3-\sqrt{2}$$