# How do you use the Binomial Theorem to expand (x+1+x^-1)^4?

Jan 17, 2016

See explanation...

#### Explanation:

This is a multiplication of trinomials not binomials, so the Binomial Theorem does not help much. You could use it by multiplying by ${\left(x - 1\right)}^{4}$ then dividing...

The Binomial Theorem allows you to find that:

${\left(a + b\right)}^{4} = {\sum}_{k = 0}^{4} \left(\begin{matrix}4 \\ k\end{matrix}\right) {a}^{4 - k} {b}^{k}$

$= {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$

We can turn our trinomial into a binomial temporarily as follows:

$\left(x - 1\right) \left(x + 1 + {x}^{- 1}\right) = \left({x}^{2} - {x}^{- 1}\right)$

Then:

${\left(x - 1\right)}^{4} {\left(x + 1 + {x}^{- 1}\right)}^{4} = {\left({x}^{2} - {x}^{- 1}\right)}^{4}$

$= {x}^{8} - 4 {x}^{5} + 6 {x}^{2} - 4 {x}^{- 1} + {x}^{- 4}$

And:

${\left(x - 1\right)}^{4} = {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$

Then long divide coefficients, including $0$'s for the missing powers of $x$ in the dividend... ${\left(x + 1 + {x}^{- 1}\right)}^{4}$

$= {x}^{4} + 4 {x}^{3} + 10 {x}^{2} + 16 x + 19 + 16 {x}^{- 1} + 10 {x}^{- 2} + 4 {x}^{- 3} + {x}^{- 4}$

An easier approach is to simply long multiply coefficients three times: ${\left(x + 1 + {x}^{- 1}\right)}^{4}$

$= {x}^{4} + 4 {x}^{3} + 10 {x}^{2} + 16 x + 19 + 16 {x}^{- 1} + 10 {x}^{- 2} + 4 {x}^{- 3} + {x}^{- 4}$

Trinomial Triangle

Another way of expressing the calculation of these coefficients is to construct a "Trinomial Triangle", similar to Pascal's triangle, but with slightly different rules: Each number is the sum of the three numbers above it: left, centre and right.

This is not quite as interesting and general in its use as Pascal's triangle, but it does specifically work well for ${\left(x + 1 + {x}^{- 1}\right)}^{n}$