1. **Problem Statement:** Find a cubic equation with real coefficients given that two of its roots are $1$ and $3+2i$. Also, state the result being used. 2. **Key Result Used:** If a polynomial has real coefficients and a complex root $a+bi$ (where $b \neq 0$), then its conjugate $a-bi$ is also a root. 3. **Identify Roots:** Given roots are $1$ and $3+2i$. By the conjugate root theorem, the third root must be $3-2i$. 4. **Form the Polynomial:** The cubic polynomial with roots $r_1, r_2, r_3$ is given by $$ (x - r_1)(x - r_2)(x - r_3) = 0 $$ Substitute roots: $$ (x - 1)(x - (3+2i))(x - (3-2i)) = 0 $$ 5. **Simplify the Complex Factors:** Multiply the complex conjugate factors: $$ (x - (3+2i))(x - (3-2i)) = \left(x - 3 - 2i\right)\left(x - 3 + 2i\right) = \left(x - 3\right)^2 - (2i)^2 = (x - 3)^2 - (-4) = (x - 3)^2 + 4 $$ 6. **Expand:** $$ (x - 3)^2 + 4 = (x^2 - 6x + 9) + 4 = x^2 - 6x + 13 $$ 7. **Final Polynomial:** $$ (x - 1)(x^2 - 6x + 13) = 0 $$ Expand: $$ x(x^2 - 6x + 13) - 1(x^2 - 6x + 13) = x^3 - 6x^2 + 13x - x^2 + 6x - 13 = x^3 - 7x^2 + 19x - 13 $$ 8. **Answer:** The cubic equation with real coefficients is $$ \boxed{x^3 - 7x^2 + 19x - 13 = 0} $$