1. **State the problem:** Solve the equation $$0 = 2x - 32x^{-3}$$ for $x$. 2. **Rewrite the equation:** The equation is $$0 = 2x - 32x^{-3}$$. 3. **Isolate terms:** Move all terms to one side: $$2x = 32x^{-3}$$ 4. **Multiply both sides by $x^3$ to eliminate negative exponent:** $$2x \cdot x^3 = 32x^{-3} \cdot x^3$$ $$2x^{4} = 32$$ 5. **Simplify:** $$2x^{4} = 32$$ 6. **Divide both sides by 2:** $$\frac{\cancel{2}x^{4}}{\cancel{2}} = \frac{32}{2}$$ $$x^{4} = 16$$ 7. **Solve for $x$ by taking the fourth root:** $$x = \pm \sqrt[4]{16}$$ 8. **Calculate the fourth root of 16:** Since $16 = 2^4$, $$x = \pm 2$$ **Final answer:** $$x = 2 \text{ or } x = -2$$