not a question, i just noticed is all. its about cube roots

Lets just proove this TheJonyMyster

$$\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\
RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\
RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\
RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\
RHS=x^{(1/n)}\\\\
RHS=\sqrt[n]{x}\\\\
RHS=LHS$$    

It's true as long as x is greater than zero.

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Yes TheJonyMyster, it is cool how roots work :)

yeah, positive x cause of imaginary numbers. also it works for any $${\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}} = {\sqrt[{{\mathtt{\left({\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}}]{{\frac{{\mathtt{x}}}{{\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}}}}}}$$

Nice proof,  Melody......!!!!

thanks Chris :)