3
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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 $256 = 2^{8}$. $\endgroup$ – Seewoo Lee Jan 9 at 2:02
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Doug Fir Jan 9 at 2:09
  • 2
    $\begingroup$ @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 $\dfrac{\sqrt{3}}{\sqrt{3}}$ $\endgroup$ – Key Flex Jan 9 at 2:11
  • $\begingroup$ This last step is called "rationalizing the denominator". $\endgroup$ – NickD Mar 2 at 17:23
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$ 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.

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

$\endgroup$
  • 1
    $\begingroup$ Thanks Marvin, this is helpful $\endgroup$ – Doug Fir Jan 9 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.