How do you expand #(1+x+x^2)^3 # using the binomial theorem?

1 Answer
Jun 1, 2016

Answer:

#(1+x+x^2)^3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6#

Explanation:

I don't think you do use the binomial theorem for this one, since #(1+x+x^2)# is a trinomial, not a binomial.

We can long multiply the coefficients a couple of times like this:

#1color(white)(00)1color(white)(00)1#
#color(white)(000)1color(white)(00)1color(white)(00)1#
#underline(color(white)(000000)1color(white)(00)1color(white)(00)1)#
#1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1#

#1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1#
#color(white)(000)1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1#
#underline(color(white)(000000)1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1)#
#1color(white)(00)3color(white)(00)6color(white)(00)7color(white)(00)6color(white)(00)3color(white)(00)1#

So:

#(1+x+x^2)^3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6#

Note that this is like picking the #n#th row of a variant of Pascal's triangle in which each number is the sum of three numbers above it:

#color(white)(0000000000000)1#
#color(white)(0000000000)1color(white)(00)1color(white)(00)1#
#color(white)(0000000)1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1#
#color(white)(0000)1color(white)(00)3color(white)(00)6color(white)(00)7color(white)(00)6color(white)(00)3color(white)(00)1#