<s>[INST] Simplify $\sqrt{\dfrac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$ into $\dfrac{\sqrt{3}}{3}$ I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$$
The solution is $\displaystyle \frac{\sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $\frac{\sqrt[3]{64}}{\sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $\dfrac{\sqrt{3}}{3}$
 [/INST] $ 256 $ is a perfect square because $ 16 \cdot 16 = 16^2 = 256 $. Thus $ \sqrt{256} = 16 $. The expression $$ \sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}} $$ evaluates to $$ \sqrt{\frac{4 + 4}{8 + 16}} $$ which simplifies to $$ \sqrt{\frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ \sqrt{\frac{8/8}{24/8}} = \sqrt{\frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} $. Using this property,
$$ \sqrt{\frac{1}{3}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ \sqrt{3} $ to achieve the desired result.
$$ \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} =  \frac{\sqrt{3}}{3} $$
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