1. **Stating the problem:** We want to decompose the rational expression $$\frac{x^2 - 1}{x^3 + x + 1}$$ into partial fractions. 2. **Check the degree:** The numerator degree is 2 and the denominator degree is 3, so partial fraction decomposition is possible without polynomial division. 3. **Factor the denominator if possible:** The denominator is $$x^3 + x + 1$$. This cubic does not factor easily with rational roots, so we treat it as an irreducible cubic. 4. **Set up the partial fraction form:** Since the denominator is an irreducible cubic, the partial fraction decomposition will be: $$\frac{x^2 - 1}{x^3 + x + 1} = \frac{Ax^2 + Bx + C}{x^3 + x + 1}$$ 5. **Because the denominator is not factorable, the expression is already in simplest form.** 6. **Conclusion:** The given expression cannot be decomposed further into simpler partial fractions with linear or quadratic denominators. **Final answer:** $$\frac{x^2 - 1}{x^3 + x + 1}$$ is already in its simplest form for partial fractions.