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. **Test $a=-4$:** $$(-4)^3 + (-4)^2 = -64 + 16 = -48 eq 36$$ 8. **Check for other roots:** The equation is cubic, so there may be other roots. Use the substitution $a = x$ and solve: $$a^3 + a^2 - 36 = 0$$ 9. **Use the Rational Root Theorem:** Possible rational roots are factors of 36: $\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$. 10. **Test $a=-3$:** $$(-3)^3 + (-3)^2 = -27 + 9 = -18 eq 36$$ 11. **Since $a=3$ is a root, divide the polynomial by $(a-3)$:** $$\frac{a^3 + a^2 - 36}{a - 3} = a^2 + 4a + 12$$ 12. **Solve the quadratic:** $$a^2 + 4a + 12 = 0$$ 13. **Use quadratic formula:** $$a = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 12}}{2} = \frac{-4 \pm \sqrt{16 - 48}}{2} = \frac{-4 \pm \sqrt{-32}}{2}$$ 14. **Simplify the discriminant:** $$\sqrt{-32} = \sqrt{-1 \times 16 \times 2} = 4i\sqrt{2}$$ 15. **Final solutions for quadratic:** $$a = \frac{-4 \pm 4i\sqrt{2}}{2} = -2 \pm 2i\sqrt{2}$$ 16. **Summary of solutions:** - Real root: $$a = 3$$ - Complex roots: $$a = -2 + 2i\sqrt{2}$$ and $$a = -2 - 2i\sqrt{2}$$ **Final answer:** $$\boxed{a = 3, \quad a = -2 \pm 2i\sqrt{2}}$$