The last term in the expansion of the binomial (3x+6)^3 is?

1. Use Cube of sum : (a+b)^3 = a^3+3a^2b+3ab^2+b^3
(3x+6)((3x)^2+2*3x*6+6^2)

2. Use multiplication distributive property (xy)^a=x^ay^a
(3x+6)(3^2x^2+2*3x+6+6^2)

3. Simplify 3^2 to 9
(3x+6)(9X^2+2*3X*6+6^2)

4. Simplify 6^2 to 36
(3x+6)(9x^2+2*3x*6+36)

5. Simplify 2*3x*6 to 36x
(3x+6)(9x^2+36x+36)

6. Distribute sum groups
3x(9x^2+36x+36)+6(9x^2+36x+36)

7. Distribute
27x^3+108x^2+108x+6(9x^2+36x+36)

8. Distribute 
27x^3+108x^2+108x+54x^2+216x+216

9. Collect like terms
27x^3+(108x^2+54x^2)+(108x+216x)+216

10. Simplify 
27x^3+162x^2+324x+216

The last term is therefore, 216. Have a nice day :D