1. **State the problem:** We need to find the range of values of $a$ (where $a \neq 0$) such that the cubic equation $$2x^3 - 3ax^2 + a^3 - 2a = 0$$ has exactly one real root. 2. **Recall the rule for roots of cubic equations:** A cubic equation can have either one real root and two complex conjugate roots, or three real roots (some possibly repeated). 3. **Use the discriminant of a cubic:** For a cubic equation $$ax^3 + bx^2 + cx + d = 0,$$ the discriminant $\Delta$ determines the nature of roots: - If $\Delta > 0$, three distinct real roots. - If $\Delta = 0$, multiple roots (at least two equal real roots). - If $\Delta < 0$, one real root and two complex conjugates. 4. **Rewrite the given equation:** $$2x^3 - 3ax^2 + 0 \cdot x + (a^3 - 2a) = 0,$$ so coefficients are: $$A=2, B=-3a, C=0, D=a^3 - 2a.$$ 5. **Calculate the discriminant $\Delta$ for the cubic:** $$\Delta = B^2C^2 - 4AC^3 - 4B^3D - 27A^2D^2 + 18ABCD.$$ Since $C=0$, terms with $C$ vanish: $$\Delta = -4B^3D - 27A^2D^2.$$ 6. **Substitute values:** $$B = -3a, \quad D = a^3 - 2a, \quad A=2.$$ Calculate: $$B^3 = (-3a)^3 = -27a^3,$$ so $$-4B^3D = -4(-27a^3)(a^3 - 2a) = 108a^3(a^3 - 2a) = 108a^6 - 216a^4,$$ and $$-27A^2D^2 = -27 \times 4 \times (a^3 - 2a)^2 = -108(a^3 - 2a)^2.$$ 7. **Expand $(a^3 - 2a)^2$:** $$ (a^3 - 2a)^2 = a^6 - 4a^4 + 4a^2.$$ 8. **Substitute back:** $$\Delta = 108a^6 - 216a^4 - 108(a^6 - 4a^4 + 4a^2) = 108a^6 - 216a^4 - 108a^6 + 432a^4 - 432a^2.$$ Simplify: $$\Delta = (108a^6 - 108a^6) + (-216a^4 + 432a^4) - 432a^2 = 216a^4 - 432a^2.$$ 9. **Factor out common terms:** $$\Delta = 216a^2(a^2 - 2).$$ 10. **Analyze the sign of $\Delta$ for $a \neq 0$:** - Since $216a^2 > 0$ for all $a \neq 0$, the sign of $\Delta$ depends on $(a^2 - 2)$. - If $a^2 - 2 > 0 \implies a^2 > 2 \implies |a| > \sqrt{2}$, then $\Delta > 0$ (three distinct real roots). - If $a^2 - 2 = 0 \implies |a| = \sqrt{2}$, then $\Delta = 0$ (multiple roots). - If $a^2 - 2 < 0 \implies |a| < \sqrt{2}$, then $\Delta < 0$ (one real root only). 11. **Conclusion:** The equation has exactly one real root only when $$|a| < \sqrt{2}$$ with $a \neq 0$. **Final answer:** $$\boxed{a \in (-\sqrt{2}, 0) \cup (0, \sqrt{2})}.$$