3√x=√x3√x
it solves out fine manually and all its just interesting.
yeah, positive x cause of imaginary numbers. also it works for any n√x=(n−1)√xn√x
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