How do you use the Binomial Theorem to expand (1+x+x^2)^3?

Nov 4, 2016

Use a variant of Pascal's triangle to find:

${\left(1 + x + {x}^{2}\right)}^{3} = 1 + 3 x + 6 {x}^{2} + 7 {x}^{3} + 6 {x}^{4} + 3 {x}^{5} + {x}^{6}$

Explanation:

This is a power of a trinomial, not a binomial so the binomial theorem does not help much.

However, note that $1 , x , {x}^{2}$ are in geometric progression, so we can use a variant of Pascal's triangle to find the coefficients we want. Each term in this variant is the sum of the three terms above it, rather than two...

$\textcolor{w h i t e}{0000000000} 1$

$\textcolor{w h i t e}{0000000} 1 \textcolor{w h i t e}{00} 1 \textcolor{w h i t e}{00} 1$

$\textcolor{w h i t e}{0000} 1 \textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{00} 1$

$\textcolor{w h i t e}{0} 1 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00} 3 \textcolor{w h i t e}{00} 1$

Hence we find:

${\left(1 + x + {x}^{2}\right)}^{3} = 1 + 3 x + 6 {x}^{2} + 7 {x}^{3} + 6 {x}^{4} + 3 {x}^{5} + {x}^{6}$