1. **State the problem:** Solve the equation $$p^{\frac{2}{3}} - 2p^{\frac{1}{3}} = 8$$ for $p$. 2. **Substitute:** Let $x = p^{\frac{1}{3}}$. Then $p^{\frac{2}{3}} = (p^{\frac{1}{3}})^2 = x^2$. 3. **Rewrite the equation:** Using substitution, the equation becomes: $$x^2 - 2x = 8$$ 4. **Bring all terms to one side:** $$x^2 - 2x - 8 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$ 6. **Calculate the roots:** - $x = \frac{2 + 6}{2} = \frac{8}{2} = 4$ - $x = \frac{2 - 6}{2} = \frac{-4}{2} = -2$ 7. **Recall substitution:** $x = p^{\frac{1}{3}}$, so $$p^{\frac{1}{3}} = 4 \quad \text{or} \quad p^{\frac{1}{3}} = -2$$ 8. **Solve for $p$:** Cube both sides: - If $p^{\frac{1}{3}} = 4$, then $p = 4^3 = 64$ - If $p^{\frac{1}{3}} = -2$, then $p = (-2)^3 = -8$ 9. **Final answer:** $$p = 64 \quad \text{or} \quad p = -8$$