Solve for x

Here is my solution. 

^{$x^\frac{1}{3} = y$}

y = 4 / y 

y^2 = 4

y = 2

$x^\frac{1}{3}=2$

x = the cube root of 2

I hope this helped,

Gavin

Never mind, I though the denominator was $x^\frac{1}{3}$

My answer is wrong. 

Solve for x:
x^(1/3) = 4/x^3

Raise both sides to the power of three:
x = 64/x^9

Cross multiply:
x^10 = 64        Take the 10th root of both sides
x = 2^(3/5)      or        x= (-2)^(3/5)               

These 2 solutions are the only real solution to balance the equation.

divide maybe?

x^{1 \over3}=4{1 \over x^3}

$x^{1 \over3}=4{1 \over x^3}\\ x^{1 \over3}=4x^{-3}\\ x=4^3*x^{-9}\\ x^{10}=4^3\\ x=4^{3/10}\\ x=2^{3/5}\\ x=\sqrt[5]{8}\\ x\approx 1.5157$

this is the only positive real answer. 

I think our guest's negative answer also works.  It may not work with convention though

The cube root is in the question. so is a negative allowed.. I am not sure.

I think there are other complex answers as well.

One more try:

a$x^{\frac{1}{3}}=4{\frac{1}{x^3}}$

definition:

${\color{black}is}\ 4\frac{1}{3}=4 +\frac{1}{3}\ {\color{black}so\ is}\ 4\frac{1}{x^3}=4 +\frac{1}{x^3}$

$x^{\frac{1}{3}}=4+{\frac{1}{x^3}}$         |  $\times x^3$

$x^{\frac{10}{3}}=4x^3+1$

$f(x)=4x^3-x^{\frac{10}{3}}+1 \neq 0$

The function has a minimum in P (0;1).
The function has no zero.

  !