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I need to find $[x^3](1-x)^{-1}(1-2x)^6$, where $[x^3]$ means the coefficent of the $[x^3]$ term. here's what I've done:

$[x^3](1-x)^{-1}(1-2x)^6=[x^3](\sum_{k=0}^6 {6\choose k}(-2x)^k)(\sum_{m=0}^\infty {m\choose 0}x^m)$

$= \sum_{k=0}^6 {6\choose k}(-2)^k[x^{3-k} ](\sum_{m=0}^\infty {m\choose 0}x^m)$

$= \sum_{k=0}^3 {6\choose k}(-2)^k[x^{3-k} ](\sum_{m=0}^\infty {m\choose 0}x^m)$ since we need $3-k \geq 0$

$= \sum_{k=0}^3 ({6\choose k}(-2)^k {3-k\choose 0})$

$= \sum_{k=0}^3 ({6\choose k}(-2)^k)$

$= {6\choose0} + (-2){6\choose1} + (4){6\choose2} + (-8){6\choose3}$

$=1-12+60-160$

$= -111$

But when I do the expansion on WolframAlpha, I see that $[x^0]=1$, $[x^1]=-12$, $[x^3]=-160$, so what am I doing wrong?

(I am following a similar idea to Trevor Gunn's answer in this question In how many ways the sum of 5 thrown dice is 25?)

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  • $\begingroup$ Did you confuse $(-x)^k$ and $x^{-k}$? Also, did you mean $x^3$ where you wrote $x^4$? $\endgroup$ – J. W. Tanner May 22 at 0:29
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    $\begingroup$ I might have worded the question confusingly, the $[x^3]$ is not multiplication, but rather finding the coefficient of the $x^3$ term in the equation that follows $\endgroup$ – Mark Dodds May 22 at 0:36
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    $\begingroup$ Oh, then maybe you should say finding the coefficient of $x^3$ in ... $\endgroup$ – J. W. Tanner May 22 at 0:37
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    $\begingroup$ Your work is correct, and this WolframAlpha link verifies it. What specifically did you see that made you think you were wrong? $\endgroup$ – Mike Earnest May 22 at 0:55
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    $\begingroup$ @JohnOmielan looking back i think that is what has happened. Thanks for clearing that up $\endgroup$ – Mark Dodds May 22 at 0:59
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As Mike Earnest confirmed in the comments, your work is correct. As I commented, and you've stated it's likely the case, the WolframAlpha results of $[x^0]=1$, $[x^1]=-12$, $[x^3]=-160$ probably come from the coefficients in the power expansion of $(1-2x)^6$ instead. You can see this directly from the first, second & fourth terms in your second last highlighted line, i.e.,

$=1-12+60-160$

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