Solve for \( x \) 4. \( x^{\frac{4}{3}}-\sqrt[3]{x^{2}}-72=0 \)

To solve the equation \( x^{\frac{4}{3}} - \sqrt[3]{x^2} - 72 = 0 \), let's first rewrite it in a more manageable form. We can substitute \( y = \sqrt[3]{x} \), which implies that \( x = y^3 \). Therefore, we have: \[ x^{\frac{4}{3}} = (y^3)^{\frac{4}{3}} = y^4 \quad \text{and} \quad \sqrt[3]{x^2} = \sqrt[3]{(y^3)^2} = y^2. \] Now substitute these into the original equation: \[ y^4 - y^2 - 72 = 0. \] Next, let’s set \( z = y^2 \), so \( y^4 = z^2 \). This changes the equation to: \[ z^2 - z - 72 = 0. \] Now we can solve this quadratic equation using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here \( a = 1, b = -1, c = -72 \): \[ z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 288}}{2} = \frac{1 \pm \sqrt{289}}{2} = \frac{1 \pm 17}{2}. \] This gives us two possible solutions for \( z \): \[ z = \frac{18}{2} = 9 \quad \text{and} \quad z = \frac{-16}{2} = -8. \] Since \( z \) represents \( y^2 \) and must be non-negative, we discard \( z = -8 \) and keep \( z = 9 \). Now reverting back to \( y^2 = 9 \), we find: \[ y = 3 \quad \text{or} \quad y = -3. \] Since \( y = \sqrt[3]{x} \), we take \( y = 3 \) (as we are looking for real \( x \)): \[ \sqrt[3]{x} = 3 \implies x = 3^3 = 27. \] Thus, the solution for \( x \) is: \[ \boxed{27}. \]