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:** Move 36 to the other side: $$a^2(a + 1) = 36$$ 5. **Try possible integer roots:** Since 36 is a perfect square and cube, test integer values for $$a$$. 6. **Test $$a=2$$:** $$2^3 + 2^2 = 8 + 4 = 12 \neq 36$$ 7. **Test $$a=3$$:** $$3^3 + 3^2 = 27 + 9 = 36$$ which satisfies the equation. 8. **Test $$a=-4$$:** $$(-4)^3 + (-4)^2 = -64 + 16 = -48 \neq 36$$ 9. **Conclusion:** The solution is $$a = 3$$. **Final answer:** $$\boxed{3}$$