Question

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

Answer 1

`(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`