# Why is it different?

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

Note by Bile Carlos
2 years, 2 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".)

- 2 years, 2 months ago

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

- 2 years, 2 months ago

There is no difference in the above expressions

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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. $$\smile$$

- 2 years, 2 months ago

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

- 2 years, 2 months ago