1. **State the problem:** Find the sum of the polynomials $$\left(\frac{1}{2} x^3 + 3x^2 - \frac{1}{4}\right) + \left(-\frac{1}{3} x^2 - 2x + 1\right).$$ 2. **Write the expression:** $$\frac{1}{2} x^3 + 3x^2 - \frac{1}{4} - \frac{1}{3} x^2 - 2x + 1$$ 3. **Group like terms:** $$\left(\frac{1}{2} x^3\right) + \left(3x^2 - \frac{1}{3} x^2\right) + \left(-2x\right) + \left(-\frac{1}{4} + 1\right)$$ 4. **Simplify each group:** - For $x^3$ term: $\frac{1}{2} x^3$ - For $x^2$ terms: $3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}$, so $\frac{8}{3} x^2$ - For $x$ term: $-2x$ - For constants: $-\frac{1}{4} + 1 = -\frac{1}{4} + \frac{4}{4} = \frac{3}{4}$ 5. **Write the final sum:** $$\frac{1}{2} x^3 + \frac{8}{3} x^2 - 2x + \frac{3}{4}$$ 6. **Conclusion:** The correct sum matches the option: $$\frac{1}{2} x^3 + \frac{8}{3} x^2 - 2x + \frac{3}{4}.$$