1. **State the problem:** Solve the equation $$a^3 + a^2 = 36$$ for the variable $a$. 2. **Rewrite the equation:** We want to find $a$ such that $$a^3 + a^2 - 36 = 0$$. 3. **Factor the equation if possible:** Factor out the common term $a^2$ from the first two terms: $$a^2(a + 1) - 36 = 0$$ 4. **Isolate terms:** Rewrite as $$a^2(a + 1) = 36$$ 5. **Try possible integer roots:** Since 36 is a perfect square and cube, test integer values for $a$. 6. **Test $a=3$:** $$3^3 + 3^2 = 27 + 9 = 36$$ which satisfies the equation. 7. **Check for other roots:** The equation is cubic, so there may be other roots. Use the original form: $$a^3 + a^2 - 36 = 0$$ 8. **Use polynomial division to factor out $(a-3)$:** Divide $$a^3 + a^2 - 36$$ by $(a-3)$: $$\frac{a^3 + a^2 - 36}{a - 3} = a^2 + 4a + 12$$ 9. **Solve quadratic $a^2 + 4a + 12 = 0$:** Use quadratic formula: $$a = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 12}}{2} = \frac{-4 \pm \sqrt{16 - 48}}{2} = \frac{-4 \pm \sqrt{-32}}{2}$$ 10. **Simplify the discriminant:** $$\sqrt{-32} = \sqrt{32}i = 4\sqrt{2}i$$ 11. **Final roots from quadratic:** $$a = \frac{-4 \pm 4\sqrt{2}i}{2} = -2 \pm 2\sqrt{2}i$$ 12. **Summary of solutions:** - Real root: $$a = 3$$ - Complex roots: $$a = -2 + 2\sqrt{2}i$$ and $$a = -2 - 2\sqrt{2}i$$ **Final answer:** $$\boxed{a = 3, \quad a = -2 \pm 2\sqrt{2}i}$$