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:** 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. **Check for possible integer roots:** Since $36$ is a perfect square and cube, try integer values for $a$. 6. **Test $a=3$:** $$3^3 + 3^2 = 27 + 9 = 36$$ which satisfies the equation. 7. **Test $a=-4$:** $$(-4)^3 + (-4)^2 = -64 + 16 = -48 eq 36$$ 8. **Test $a=-1$:** $$(-1)^3 + (-1)^2 = -1 + 1 = 0 eq 36$$ 9. **Conclusion:** The only real solution is $$a = 3$$. **Final answer:** $$a = 3$$