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