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

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

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Answer 1 · 2016-11-04T00:44:42

Use a variant of Pascal's triangle to find:

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

Explanation:

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

However, note that $1, x, x^{2}$ $1, x, x^2$ 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...

$00000000001$ $color(white)(0000000000)1$

$00000001001001$ $color(white)(0000000)1color(white)(00)1color(white)(00)1$

$00001002003002001$ $color(white)(0000)1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1$

$01003006007006003001$ $color(white)(0)1color(white)(00)3color(white)(00)6color(white)(00)7color(white)(00)6color(white)(00)3color(white)(00)1$

Hence we find:

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

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

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