1. **State the problem:** Solve the equation $$2x^{\frac{2}{3}} - x^{\frac{1}{3}} = 3$$ for $x$. 2. **Introduce substitution:** Let $y = x^{\frac{1}{3}}$. Then $x^{\frac{2}{3}} = y^2$. 3. **Rewrite the equation:** Substitute into the original equation: $$2y^2 - y = 3$$ 4. **Rearrange into standard quadratic form:** $$2y^2 - y - 3 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-1$, and $c=-3$. Calculate the discriminant: $$\Delta = (-1)^2 - 4 \times 2 \times (-3) = 1 + 24 = 25$$ So, $$y = \frac{1 \pm 5}{4}$$ 6. **Find the two solutions for $y$:** - $$y_1 = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}$$ - $$y_2 = \frac{1 - 5}{4} = \frac{-4}{4} = -1$$ 7. **Back-substitute to find $x$:** Recall $y = x^{\frac{1}{3}}$, so $$x = y^3$$ Calculate each: - $$x_1 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$ - $$x_2 = (-1)^3 = -1$$ 8. **Final answer:** $$x = \frac{27}{8}, -1$$ These are the simplified real solutions to the equation.