Solve the Nested Radical: Simplifying ⁶√(√2)

Let's use the definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

$\sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}$

Let's return to the problem and convert using the root definition we mentioned above the roots in the problem:

$\sqrt[6]{\sqrt{2}}=\sqrt[6]{2^{\frac{1}{2}}}=\big(2^{\frac{1}{2}}\big)^{\frac{1}{6}}$

where in the first stage we applied the root definition as a power mentioned earlier to the inner expression (meaning inside the larger-outer root) and then we used parentheses and applied the same definition to the outer root.

Now let's remember the power law for power of a power:

$(a^m)^n=a^{m\cdot n}$

Let's apply this law to the expression we got in the last stage:

$\big(2^{\frac{1}{2}}\big)^{\frac{1}{6}}=2^{\frac{1}{2}\cdot\frac{1}{6}}=2^{\frac{1\cdot1}{2\cdot6}}=2^{\frac{1}{12}}$

where in the first stage we applied the power law mentioned above and then simplified the resulting expression and performed the multiplication of fractions in the power exponent.

Let's summarize the solution steps so far, we got that:

$\sqrt[6]{\sqrt{2}}=\big(2^{\frac{1}{2}}\big)^{\frac{1}{6}} =2^{\frac{1}{12}}$

In the next stage we'll apply again the root definition as a power that was mentioned at the beginning of the solution, but in the opposite direction:

$a^{\frac{1}{n}} = \sqrt[n]{a}$

Let's apply this law to go back and present the expression we got in the last stage in root form:

$$$2^{\frac{1}{12}} =\sqrt[12]{2}$

Therefore we got that:

$\sqrt[6]{\sqrt{2}}=2^{\frac{1}{12}} =\sqrt[12]{2}$

Therefore the correct answer is answer A.