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Why is it different?

Why is \( \sqrt[3]{x^2} \) different from \(x^{2/3} \)?

Note by Bile Carlos
1 year, 6 months ago

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For real numbers they are equivalent. In the complex case, \(\sqrt[3]{i^2}\) usually means the real third root of \(-1\), whereas \(i^\frac{2}{3}\) usually refers to the \(principal\) third root of \(-1\) (it has three third roots), which is \(\frac{1}{2} + \frac{\sqrt{3}}{2} i\). For instance, you'll get these different values if you enter the expressions in Google or Wolfram Alpha. Enter "cube root of i^2" and you get -1; enter "i^(2/3)" and you get 0.5 + 0.866025404 i. This is purely conventional. (Wolfram helpfully tells you, "Assuming 'cube root' is the real-valued root".)

Mark C - 1 year, 6 months ago

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Yep but, the domain of third root of x^2 is different from the domain of x^(2/3). How is that possible??

Bile Carlos - 1 year, 6 months ago

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There is no difference in the above expressions

Sai B - 1 year, 6 months ago

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Yep but, the domain of third root of x^2 is different from the domain of x^(2/3). How is that possible??

Bile Carlos - 1 year, 6 months ago

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There is no difference in the two expressions. But since you are asking then:

\(\sqrt[3]{x^2}=x^{\dfrac{2}{3}}\)

But

\(\large \sqrt{x^{\sqrt[3]{2}}}=(x)^{\dfrac{2^{\frac{1}{3}}}{2}}=(x)^{{2^{-\frac{2}{3}}}}\)

Hope this helps. \(\smile\)

Abhay Tiwari - 1 year, 6 months ago

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Yep but, the domain of third root of x^2 is different from the domain of x^(2/3). How is that possible??

Bile Carlos - 1 year, 6 months ago

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