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?
On
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.
On
$\sqrt[3]{x^3}$ 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.
But you're wrong: For $x\in \mathbb R$, $$\sqrt[3]{x^3} = x$$ plain and simple.
Take a simple example: $$\sqrt[3]{(-1)^3} = -1 \neq |-1| = 1$$