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

The last term will be the second part cubed or 6^3 which equals 216

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