Finding the Coefficient of x^6y^3 using Binomial Theorem in (3x-2y)^9

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Homework Help Overview

The discussion revolves around finding the coefficient of \(x^6y^3\) in the expression \((3x-2y)^9\) using the binomial theorem. Participants are exploring the application of the theorem and the necessary calculations involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the binomial theorem by expanding the expression fully to identify the coefficient. Some participants suggest that a full expansion is unnecessary and that focusing on specific terms would suffice.

Discussion Status

There are mixed opinions regarding the approach to the problem. Some participants provide guidance on using the binomial theorem more effectively, while others question the need for a complete polynomial expansion. The conversation reflects a productive exploration of different methods to arrive at the desired coefficient.

Contextual Notes

Participants discuss the implications of using the binomial theorem correctly and the importance of identifying specific terms rather than expanding the entire expression. There is also mention of verifying results with external tools, indicating a reliance on computational resources for confirmation.

reenmachine
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Homework Statement



Use the binomial theorem to find the coefficient of ##x^6y^3## in ##(3x-2y)^9##.

Homework Equations



##1+9+36+84+126+126+84+36+9+1##

(I used two lines for the lenght)

##1(3x)^9(-2y)^0+9(3x)^8(-2y)^1+36(3x)^7(-2y)^2+84(3x)^6(-2y)^3+126(3x)^5(-2y)^4##
##+126(3x)^4(-2y)^5+84(3x)^3(-2y)^6+36(3x)^2(-2y)^7+9(3x)^1(-2y)^8+1(3x)^0(-2y)^9##

(again using two lines because of the lenght)

##19683x^9 - 118098x^8y + 314928x^7y^2 -489888x^6y^3 +489888x^5y^4 -326592x^4y^5##
##+145152x^3y^6 - 41472x^2y^7 +6912xy^8 -512y^9##

So the coefficient of ##x^6y^3## would be ##-489888##

any help will be greatly appreciated! thank you!
 
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You can check your result with WolframAlpha.
I don't see where you would need help.
 
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mfb said:
You can check your result with WolframAlpha.
I don't see where you would need help.

thank you! I wasn't aware of that site
 
There's really no need to write out the full polynomial. You only need one term.
 
micromass said:
There's really no need to write out the full polynomial. You only need one term.

You're right , I guess since it was the first time I used the theorem without a perfect ##(x+y)^n## I wanted to verify it.

thank you!
 
The problem, according to you, said "Use the binomial theorem" and you did NOT do that.

The binomial theorem says that (a+ b)^n= \sum_{i=0}^\infty \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}. Here a= 3x and b= -2y. You want "the coefficient of x6y3 with n= 9.

So this is the "i= 6" term: \begin{pmatrix}9 \\ 6\end{pmatrix}(3x)^6(-2y)^3= \begin{pmatrix}9 \\ 6\end{pmatrix}(729)x^6(-8)y^3 so the coefficient is \begin{pmatrix}9 \\ 6\end{pmatrix}(729)(-8)= \begin{pmatrix}9 \\ 6\end{pmatrix}= 5832\begin{pmatrix}9 \\ 6\end{pmatrix}.

Of course, \begin{pmatrix}9 \\ 6 \end{pmatrix}= \frac{9!}{6!3!}= \frac{9(8)(7)}{6}= 3(4)(7)= 84.
 
Yes, as HallsofIvy pointed out, the whole point here of using the Binomial Theorem is that you DO NOT actually have to do a full expansion. You just need to pick the appropriate terms out of the summation!
 

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