# 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 $$\dfrac{\sqrt{3}}{3}$$

• As you said, you can just simplify all the radicals as $\sqrt{64} = 8, \sqrt{256} = 16, \sqrt[3]{64}=4, \sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$. Jan 9 '19 at 2:02

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 $$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}=\sqrt{\dfrac{4+4}{8+16}}=\sqrt{\dfrac{8}{24}}=\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt{3}}{3}$$

• Your last step that goes from $\sqrt{\frac{1}{3}}$ to $\frac{\sqrt{3}}{3}$, how did you do that? What's the rule there? Jan 9 '19 at 2:09
• @DougFir $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{\sqrt{3}}\cdot\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$. I just multiplied the final answer with $\dfrac{\sqrt{3}}{\sqrt{3}}$ Jan 9 '19 at 2:11
• This last step is called "rationalizing the denominator". Mar 2 '19 at 17:23

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

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

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

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

On rationalization,

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

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

Thanks.

$$256$$ 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.

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

• Thanks Marvin, this is helpful Jan 9 '19 at 18:45