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 $$a^3 + a^2 - 36 = 0.$$ This is a cubic equation. 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 positive, try integer values for $a$ that satisfy the equation. 6. **Test $a=3$:** $$3^3 + 3^2 = 27 + 9 = 36,$$ which satisfies the equation. 7. **Check for other roots:** The cubic can have up to 3 real roots. To find others, rewrite as $$a^3 + a^2 - 36 = 0.$$ 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:** $$\sqrt{-32} = \sqrt{32}i = 4\sqrt{2}i,$$ so $$a = \frac{-4 \pm 4\sqrt{2}i}{2} = -2 \pm 2\sqrt{2}i.$$ 11. **Final solutions:** $$a = 3,$$ $$a = -2 + 2\sqrt{2}i,$$ $$a = -2 - 2\sqrt{2}i.$$ Only $a=3$ is a real solution. **Answer:** $$\boxed{3}$$