# Difficulty using binomial theorem with two sets of terms in brackets

Expand (1+x+x2)(1âˆ′x)8(1 + x + x^2)(1 âˆ’ x)^8 in ascending powers of xx, up to and including, x3x^3 using binomial theorem.

I have been trying to study this, and have been using this URL: http://www.purplemath.com/modules/binomial2.htm

Although I understand the fundamentals (not committed to memory 100% but close) for one set of brackets, I do not know how to approach the situation with 2 sets of brackets as shown above.

Please could someone explain the method for this with 2 sets of brackets in a step by step way that is easy to understand, for someone who is pretty much a beginner with this stuff. please keep it within my realm of understanding. I would be very grateful if someone could do this.

edit..

I know that if it was just (1−8)8(1-8)^8
then I could plug it into the formula same as below.

(Am i correct to say I would only go until (1−8)8−3(1-8)^{8-3} for my question ?)

But i DO NOT know how to plug it in to the formula when there are 2 sets of parentheses. my question is how to I apply the formula shown in the picture when there are two parentheses?

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2

Could you please expand this as far as you are able? Or expand one of the terms with binomial theorem.
– Larry B.
Oct 20 at 22:39

So I guess you started by using the binomial theorem to expand (1−x)8(1-x)^8 and then you multiplied the resulting expansion by 1+x+x2?1+x+x^2? At what point did you get stuck?
– bof
Oct 20 at 22:42

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1

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\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}
\begin{align}
&\pars{1 + x + x^{2}}\pars{1 âˆ’ x}^{8} =
\pars{1 – x^{3}}\pars{1 – x}^{7} =
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} –
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n + 3}
\\[5mm] = &\
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} +
\sum_{n = 3}^{10}{7 \choose n – 3}\pars{-1}^{n}x^{n}
\\[5mm] = &\
\bracks{1 – 7x + 21x^{2} + \sum_{n = 3}^{7}{7 \choose n}\pars{-1}^{n}x^{n}} +
\bracks{\sum_{n = 3}^{7}{7 \choose n – 3}\pars{-1}^{n}x^{n} + 21x^{8} – 7x^{9} + x^{10}}
\\[5mm] = &\
1 – 7x + 21x^{2} +
\sum_{n = 3}^{7}\bracks{{7 \choose n} + {7 \choose n – 3}}\pars{-1}^{n}x^{n} +
21x^{8} – 7x^{9} + x^{10} = \sum_{n = 0}^{10}a_{n}x^{n}
\end{align}\begin{align}
&\pars{1 + x + x^{2}}\pars{1 âˆ’ x}^{8} =
\pars{1 – x^{3}}\pars{1 – x}^{7} =
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} –
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n + 3}
\\[5mm] = &\
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} +
\sum_{n = 3}^{10}{7 \choose n – 3}\pars{-1}^{n}x^{n}
\\[5mm] = &\
\bracks{1 – 7x + 21x^{2} + \sum_{n = 3}^{7}{7 \choose n}\pars{-1}^{n}x^{n}} +
\bracks{\sum_{n = 3}^{7}{7 \choose n – 3}\pars{-1}^{n}x^{n} + 21x^{8} – 7x^{9} + x^{10}}
\\[5mm] = &\
1 – 7x + 21x^{2} +
\sum_{n = 3}^{7}\bracks{{7 \choose n} + {7 \choose n – 3}}\pars{-1}^{n}x^{n} +
21x^{8} – 7x^{9} + x^{10} = \sum_{n = 0}^{10}a_{n}x^{n}
\end{align}
where

a_{n} =
\left\{\begin{array}{lcl}
\ds{\phantom{-}1} & \mbox{if} & \ds{n = 0}
\\[2mm]
\ds{-7} & \mbox{if} & \ds{n = 1}
\\[2mm]
\ds{\phantom{-}21} & \mbox{if} & \ds{n = 2}
\\[2mm]
\ds{\pars{-1}^{n}\bracks{%
{7 \choose n} + {7 \choose n – 3}}} & \mbox{if} & \ds{3 \leq n \leq 7}
\\[2mm]
\ds{a_{10 -n}} & \mbox{if} & \ds{8 \leq n \leq 10}
\end{array}\right.

a_{n} =
\left\{\begin{array}{lcl}
\ds{\phantom{-}1} & \mbox{if} & \ds{n = 0}
\\[2mm]
\ds{-7} & \mbox{if} & \ds{n = 1}
\\[2mm]
\ds{\phantom{-}21} & \mbox{if} & \ds{n = 2}
\\[2mm]
\ds{\pars{-1}^{n}\bracks{%
{7 \choose n} + {7 \choose n – 3}}} & \mbox{if} & \ds{3 \leq n \leq 7}
\\[2mm]
\ds{a_{10 -n}} & \mbox{if} & \ds{8 \leq n \leq 10}
\end{array}\right.

Indeed, the coefficients \ds{\braces{a_{n}}}_{\ 0\ \leq\ n\ \leq\ 10}\ds{\braces{a_{n}}}_{\ 0\ \leq\ n\ \leq\ 10} satisfacen \ds{a_{n} = a_{10 – n}}\ds{a_{n} = a_{10 – n}} such that’s sufficient to enumerate the values of \ds{a_{n}}\ds{a_{n}} for \ds{n = 0,1,\ldots,5}\ds{n = 0,1,\ldots,5}. Namely,

a_{0} = 1\,,\ a_{1} = -7\,,\ a_{2} = 21\,, a_{3} = -36\,,\ a_{4} = 42\,,\
a_{5} = -42

a_{0} = 1\,,\ a_{1} = -7\,,\ a_{2} = 21\,, a_{3} = -36\,,\ a_{4} = 42\,,\
a_{5} = -42

I appreciate you putting time into this, please can you read my edit as it shows the method that I am beginning to understand, this topic is very confusing to me and the answers have also confused me. can you please explain in a way that I can see visually connects to the image I have added ?
– Flewitt Connor
2 days ago