Simplify $\sqrt{\dfrac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$ into $\dfrac{\sqrt{3}}{3}$

Note that $$\sqrt[3]{64}=(64)^{\frac13}=(4^3)^\frac13=4$$ $$\sqrt[4]{256}=(256)^{\frac14}=(4^4)^{\frac14}=4$$ $$\sqrt{64}=8$$ $$\sqrt{256}=16$$ So, we get $$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}=\sqrt{\dfrac{4+4}{8+16}}=\sqrt{\dfrac{8}{24}}=\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt{3}}{3}$$

You have: $$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$$ Then: $$\sqrt{\frac{\sqrt[3]{8^2} + \sqrt[4]{16^2}}{\sqrt{8^2}+\sqrt{16^2}}}$$ $$\sqrt{\frac{\sqrt[3]{2^6} + \sqrt[4]{4^4}}{8 + 16}}$$ $$\sqrt{\frac{2^2 + 4}{24}}$$ $$\sqrt{\frac{8}{8 \cdot 3}}$$ $$\frac{1}{\sqrt{3}}$$ $$\frac{\sqrt{3}}{\sqrt{3}\sqrt{3}}$$ $$\frac{\sqrt{3}}{3}$$ And this is the answer.

We know that,

$64=8×8$ or $\sqrt{64}=8$

and
$256=16×16$ or $\sqrt{256}=16$

Also, $64^\frac{1}{3}$ and $256^\frac{1}{4}$

Therefore, $64^\frac{1}{3}=(8×8)^\frac{1}{3} =(8^\frac{1}{3})×(8^\frac{1}{3})=2×2=4$

Similarly, $256^\frac{1}{4}=(16×16)^\frac{1}{4}=(16^\frac{1}{4})×(16^\frac{1}{4})=2×2=4$

Now, your expression reduces to

=$\sqrt{\frac{4+4}{8+16}}$

=$\sqrt{\frac{8}{24}}$

=$\sqrt{\frac{1}{3}}$

On rationalization,

$=\frac{1}{\sqrt{3}}×\frac{\sqrt{3}} {\sqrt{3}}$

Hence, $\frac{\sqrt{3}}{3}$

Thanks.