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 a positive integer, test integer values of $a$ that satisfy the equation. - For $a=2$: $$2^2(2+1) = 4 \times 3 = 12 \neq 36.$$ - For $a=3$: $$3^2(3+1) = 9 \times 4 = 36.$$ This satisfies the equation. 6. **Check for other roots:** The equation is cubic, so there may be other roots. Use the original equation: $$a^3 + a^2 - 36 = 0.$$ 7. **Divide the cubic by $(a-3)$ to find other roots:** Using polynomial division or synthetic division: $$\frac{a^3 + a^2 - 36}{a - 3} = a^2 + 4a + 12.$$ 8. **Solve the quadratic equation:** $$a^2 + 4a + 12 = 0.$$ Calculate the discriminant: $$\Delta = 4^2 - 4 \times 1 \times 12 = 16 - 48 = -32 < 0,$$ so no real roots here. 9. **Conclusion:** The only real solution is $$\boxed{a = 3}.$$