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 $\dfrac{\sqrt{3}}{3}$
$\endgroup$ 14 Answers
$\begingroup$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 $$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}=\sqrt{\dfrac{4+4}{8+16}}=\sqrt{\dfrac{8}{24}}=\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt{3}}{3}$$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$$\endgroup$ $\begingroup$We know that,
$64=8×8$ or $\sqrt{64}=8$
and
$256=16×16$ or $\sqrt{256}=16$Also, $64^\frac{1}{3}$ and $256^\frac{1}{4}$
Therefore, $64^\frac{1}{3}=(8×8)^\frac{1}{3} =(8^\frac{1}{3})×(8^\frac{1}{3})=2×2=4$
Similarly, $256^\frac{1}{4}=(16×16)^\frac{1}{4}=(16^\frac{1}{4})×(16^\frac{1}{4})=2×2=4$
Now, your expression reduces to
=$\sqrt{\frac{4+4}{8+16}}$
=$\sqrt{\frac{8}{24}}$
=$\sqrt{\frac{1}{3}}$
On rationalization,
$=\frac{1}{\sqrt{3}}×\frac{\sqrt{3}} {\sqrt{3}}$
Hence, $\frac{\sqrt{3}}{3}$
Thanks.
$ 256 $ is a perfect square because $ 16 \cdot 16 = 16^2 = 256 $. Thus $ \sqrt{256} = 16 $. The expression $$ \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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