1. State the problem: Solve the equation $$x^{2/3} - x^{-1/3} = 6x^{5/3}$$. 2. Multiply through by $x^{-1/3}$ to eliminate negative exponents: $$x^{2/3} \times x^{-1/3} - x^{-1/3} \times x^{-1/3} = 6x^{5/3} \times x^{-1/3}$$ Simplify each term using laws of exponents $a^m \times a^n = a^{m+n}$: $$x^{2/3 - 1/3} - x^{-1/3 - 1/3} = 6x^{5/3 - 1/3}$$ $$x^{1/3} - x^{-2/3} = 6x^{4/3}$$ 3. Introduce substitution to simplify: Let $$y = x^{1/3}$$, so: $$y - y^{-2} = 6y^{4}$$ Rewrite $y^{-2}$ as $\frac{1}{y^2}$: $$y - \frac{1}{y^{2}} = 6y^{4}$$ 4. Multiply both sides by $y^2$ to clear denominator: $$y^{3} - 1 = 6y^{6}$$ 5. Rearrange to standard polynomial form: $$6y^{6} - y^{3} + 1 = 0$$ 6. Let $$z = y^{3}$$ to reduce degree: $$6z^{2} - z + 1 = 0$$ 7. Solve quadratic equation $6z^2 - z + 1 = 0$ using quadratic formula: $$z = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 6 \times 1}}{2 \times 6} = \frac{1 \pm \sqrt{1-24}}{12} = \frac{1 \pm \sqrt{-23}}{12}$$ 8. Since discriminant $<0$, no real roots for $z$. Thus no real roots for $y^{3}$ and no real roots for $x$. 9. Conclusion: There are no real solutions to the equation. Final answer: No real solutions for $x$.