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It is known that the formula $\sqrt{x^2}$ is equal to the value of $|x|$. In my spare time last night, I wondered about $\sqrt[3]{x^3}$. After some thought and some graphing, I came up with this:

If $x<0$, $\Im(\sqrt[3]{x^3})$

If $x=0$, $0$

If $x>0$, $\Re(\sqrt[3]{x^3})$

This system is equal to $|x|$.

Why is this?

Community
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fr00ty_l00ps
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  • This isn't equal to $|x|$. $\sqrt[3]{x^3} = x$ for every $x \in \mathbb{R}$. Its imaginary part is $0$ for every real $x$. – Dan Shved Oct 28 '13 at 14:28

3 Answers3

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But you're wrong: For $x\in \mathbb R$, $$\sqrt[3]{x^3} = x$$ plain and simple.

Take a simple example: $$\sqrt[3]{(-1)^3} = -1 \neq |-1| = 1$$

amWhy
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$\sqrt[3]{x^3}$ is not equal to $|x|$. For this, consider the following examples:

  • if $x=\color{red}{+}2$ then $x^3=(\color{red}{+}2)^3=\color{red}{+}8$.

  • if $x=\color{red}{-}2$ then $x^3=(\color{red}{-}2)^3=\color{red}{-}8$.

So $$\sqrt[3]{\color{red}{+}8}=\color{red}{+}2$$ but $$\sqrt[3]{\color{red}{-}8}=\color{red}{-}2$$ We have these facts while $(\pm2)^2=4$ and so $\sqrt{4}$ is just $+2$ because $\sqrt{...}$ cannot accept any negative number inside when working with reals.

Mikasa
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Under standard conventions, $\sqrt[3]{x}=x$ for all real $x$, which you can easily show. It gets much trickier if you're working in complex numbers.

TBrendle
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