# How do you simplify $(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$ [closed]

$$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$

I want to find the coefficients of $x^{10}$

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## closed as off-topic by Zev Chonoles, O.L., Andres Caicedo, Asaf Karagila, AangJul 14 '13 at 10:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – Zev Chonoles, O.L., Asaf Karagila, Aang
If this question can be reworded to fit the rules in the help center, please edit the question.

Please indicate your thoughts. (And the current tag is baffling.) –  Did Jul 13 '13 at 8:16
What does "pls" mean? Last time I checked it was one of the two playlist formats in WinAmp. –  Asaf Karagila Jul 13 '13 at 8:37
'pls' means 'please'. –  Eckhard Jul 13 '13 at 9:40
@Eckhard Probably sarcasm ;) –  dreamer Jul 13 '13 at 10:38

The not-quite-brute-force approach is feasible here. You can rewrite the expression as

$$\frac{1-x^7}{1-x}\cdot\frac{1-x^8}{1-x}\cdot\frac{1-x^9}{1-x}\;.$$

Now

$$\frac{1-x^7}{1-x}=1+x+x^2+x^3+x^4+x^5+x^6\;.$$

and in general

$$\frac{1-x^n}{1-x}=1+x+x^2+\ldots+x^{n-1}\;.$$

$$\left(1+x+x^2+\dots+x^6\right)\left(1+x+x^2+\ldots+x^7\right)\left(1+x+x^2+\ldots+x^8\right)\;.$$

Before you combine like terms, each term in this product will have the form $x^i\cdot x^j\cdot x^k=x^{i+j+k}$, where $0\le i\le 6$, $0\le j\le 7$, and $0\le k\le 8$. Thus, you’re looking for the number of solutions to $i+j+k=10$, where $0\le i\le 6$, $0\le j\le 7$, and $0\le k\le 8$.

Finding the number of solutions in non-negative integers without any upper bounds is a standard stars-and-bars problem; accounting for the upper bounds requires a simple inclusion-exclusion argument. There are many examples of such calculation on the site; here is an answer that illustrates such a calculation.

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Corrected answer. –  Promise Jul 13 '13 at 10:06

$$(1-x^7)(1-x^8)(1-x^9)=1-x^7-x^8-x^9+\text{ powers of }x > 10$$

So, the coefficient of $x^{10}$ in $$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$

will be the coefficient of $x^{10}$ in $(1-x)^{-3}$

$-$ the coefficient of $x^{10-7}=x^3$ in $(1-x)^{-3}$

$-$ the coefficient of $x^{10-8}=x^2$ in $(1-x)^{-3}$

$-$ the coefficient of $x^{10-9}=x$ in $(1-x)^{-3}$

$$(1-x)^{-3}=1+3\cdot x+\frac{3(3+1)}{2!}x^2+\frac{3(3+1)(3+2)}{3!}x^3+\cdots +\frac{3(3+1)(3+2)\cdots(3+8)(3+9)}{10!}x^{10} +\cdots$$ if $|x|<1$

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@JonasMeyer, agreed and ammended –  lab bhattacharjee Jul 14 '13 at 9:46

Expand $(1-x)^{-3}=1+3x+6x^2+10x^3+...$. The coefficient of $x^{10}$ should now be easy to pick off.

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the idea of the question is

($1^n - b^n)= (1-b)(1 + b + b^2 + b^3...........)$

on solving, the $(1-x)$ terms in the numerator and denominator gets cancelled

and we get,

$(1 + x^2 + x^3 +....x^7)(1 + x^2 + x^3+....x^8)(1 + x^2 + x^3 +....x^9)$

now group the degrees to get $10$ $(x^10)$ :-

on finding the possible number of groups we get 60 different groups that can be formed and since the coefficient of each group is $1$,

the answer should be $60$

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Your highest exponents are off in your rewriting of the expression. The same method you suggest was correctly done in Brian M. Scott's earlier answer. –  Jonas Meyer Jul 13 '13 at 23:10