1. **State the problem:** Solve $a^3+a^2=36$ for $a$. 2. **Use a standard factoring idea:** First move everything to one side: $$a^3+a^2-36=0$$ 3. **Try to find a rational root:** Test simple whole numbers. For $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ is a solution. 4. **Factor the polynomial:** Since $a=3$ works, $(a-3)$ is a factor. Divide $a^3+a^2-36$ by $(a-3)$ to get: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 5. **Solve each factor:** $$ (a-3)(a^2+4a+12)=0 $$ So either $$a-3=0$$ or $$a^2+4a+12=0$$ 6. **Find the real solution:** $$a=3$$ For the quadratic, the discriminant is $$b^2-4ac=4^2-4(1)(12)=16-48=-32$$ Since this is negative, it has no real solutions. 7. **Final answer:** $$\boxed{a=3}$$