We are asked to solve the equation $x^{\frac{2}{3}} + x^{\frac{2}{3}} - 6 = 0$ for $x$.

1. Problem Description

We are asked to solve the equation $x^{\frac{2}{3}} + x^{\frac{2}{3}} - 6 = 0$ for $x$.

2. Solution Steps

First, simplify the equation:

$2x^{\frac{2}{3}} - 6 = 0$

Add 6 to both sides of the equation:

$2x^{\frac{2}{3}} = 6$

Divide both sides by 2:

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

Raise both sides to the power of $\frac{3}{2}$:

$(x^{\frac{2}{3}})^{\frac{3}{2}} = 3^{\frac{3}{2}}$

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

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

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

$x = 3\sqrt{3}$

Note that since we took an even root (the square root) we need to consider both positive and negative solutions.

Since $x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2$, $x^{\frac{1}{3}}$ can be positive or negative. Let $y = x^{\frac{1}{3}}$. Then $y^2 = 3$, which means $y = \pm \sqrt{3}$.

Therefore, $x^{\frac{1}{3}} = \pm \sqrt{3}$.

$x = (\pm \sqrt{3})^3 = (\pm 3^{\frac{1}{2}})^3 = \pm 3^{\frac{3}{2}} = \pm 3\sqrt{3}$.

Thus, the solutions are $x = 3\sqrt{3}$ and $x = -3\sqrt{3}$.

3. Final Answer

$x = 3\sqrt{3}, -3\sqrt{3}$