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

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

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Answer 1 · 2016-06-01T18:52:49

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

Explanation:

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

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

$1001001$ $1color(white)(00)1color(white)(00)1$
$0001001001$ $color(white)(000)1color(white)(00)1color(white)(00)1$
$underline(color(white)(000000)1color(white)(00)1color(white)(00)1)$
$1color(white)(00)2color(white)(00)3color(white)(00)2color(white)(00)1$

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

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:

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

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

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How do you expand (1+x+x^2)^3 (1+x+x2)3(1+x+x^2)^3 using the binomial theorem?

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