1. The problem is to factorize the cubic polynomial $$x^3 - 9$$. 2. Recognize that $$x^3 - 9$$ can be rewritten as $$x^3 - 3^2$$, which is a difference of cubes since $$9 = 3^2$$ is not a cube, but we can rewrite the problem as $$x^3 - 27 + 18$$ to check if it fits difference of cubes. However, the given factorization is $$(x - 3)(x^2 + 3x + 9)$$ which corresponds to the factorization of $$x^3 - 27$$. 3. Since the original polynomial is $$x^3 - 9$$, it is not a difference of cubes but a cubic minus a constant. The factorization given is for $$x^3 - 27$$. 4. To factor $$x^3 - 9$$, we can try to find roots or use polynomial division. Let's check if $$x= ext{some root}$$ satisfies the polynomial. 5. Try to find rational roots using the Rational Root Theorem. Possible roots are factors of 9: $$\pm1, \pm3, \pm9$$. 6. Evaluate $$x^3 - 9$$ at these values: - At $$x=1$$: $$1 - 9 = -8 \neq 0$$ - At $$x=3$$: $$27 - 9 = 18 \neq 0$$ - At $$x=-3$$: $$-27 - 9 = -36 \neq 0$$ - At $$x=9$$: $$729 - 9 = 720 \neq 0$$ - At $$x=-1$$: $$-1 - 9 = -10 \neq 0$$ 7. No rational roots, so $$x^3 - 9$$ is irreducible over the rationals. 8. The factorization given $$(x - 3)(x^2 + 3x + 9)$$ is for $$x^3 - 27$$, not $$x^3 - 9$$. 9. Therefore, the factorization of $$x^3 - 9$$ over real numbers is not simple; it can be expressed using cube roots or left as is. Final answer: The factorization $$(x - 3)(x^2 + 3x + 9)$$ is for $$x^3 - 27$$, not $$x^3 - 9$$. The polynomial $$x^3 - 9$$ does not factor nicely over the rationals.