1. **State the problem:** Solve for $a$ in the equation $$a^3 + a^2 = 36.$$\n\n2. **Rewrite the equation:** We want to find $a$ such that $$a^3 + a^2 - 36 = 0.$$\n\n3. **Factor the equation if possible:** Factor out the common term $a^2$ from the first two terms:\n$$a^2(a + 1) - 36 = 0.$$\n\n4. **Isolate the constant:** Move 36 to the other side:\n$$a^2(a + 1) = 36.$$\n\n5. **Try possible integer roots:** Since 36 is positive, try integer values for $a$ that satisfy the equation.\n\n6. **Test $a=3$:**\n$$3^2(3 + 1) = 9 \times 4 = 36,$$ which satisfies the equation.\n\n7. **Check for other roots:** The equation is cubic, so there may be other roots.\nRewrite as $$a^3 + a^2 - 36 = 0.$$\nUse the Rational Root Theorem to test possible roots $\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$.\n\n8. **Test $a=-4$:**\n$$(-4)^3 + (-4)^2 = -64 + 16 = -48 \neq 36.$$\n\n9. **Divide polynomial by $(a-3)$:**\nPerform polynomial division or synthetic division:\n$$\frac{a^3 + a^2 - 36}{a - 3} = a^2 + 4a + 12.$$\n\n10. **Solve quadratic $a^2 + 4a + 12 = 0$:**\nUse quadratic formula:\n$$a = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 12}}{2} = \frac{-4 \pm \sqrt{16 - 48}}{2} = \frac{-4 \pm \sqrt{-32}}{2}.$$\n\n11. **Simplify the discriminant:**\n$$\sqrt{-32} = \sqrt{32}i = 4\sqrt{2}i,$$ so\n$$a = \frac{-4 \pm 4\sqrt{2}i}{2} = -2 \pm 2\sqrt{2}i.$$\n\n12. **Final solutions:**\n$$a = 3,$$ or $$a = -2 + 2\sqrt{2}i,$$ or $$a = -2 - 2\sqrt{2}i.$$\n\nThe only real solution is $a=3$.