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Find the constants A,B,C such that $$\left(z-\frac1z\right)^5 = A\left(z^5 - \frac1{z^5}\right) + B\left(z^3 - \frac1{z^3}\right) + C\left(z-\frac1z\right)$$

I'm not sure how to find A,B,C. As far as I managed is that I think you're supposed to do binomial expansion but I'm unsure where to go after that.

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2 Answers 2

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Hint: Binomial gives $${5\choose 0}z^5-{5\choose 1}z^3+{5\choose 2}z-\frac{{5\choose 3}}z+\frac{{5\choose 4}}{z^3}-\frac{5\choose 5}{z^5}$$

Simplifying these $5\choose k$'s should give the desired result.

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  • $\begingroup$ yea saw what i did wrong. I was sloppy and completely forgot about the co-efficient parts in brackets $\endgroup$
    – John
    Nov 26, 2017 at 18:55
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we get by expanding $$z^5-\frac{1}{z^5}-5\left(z^3-\frac{1}{z^3}\right)+10\left(z-\frac{1}{z}\right)$$ can you get it from here? i have expanded $$(z-\frac{1}{z})^5=z^5-\frac{1}{z^5}-5(z^3-\frac{1}{z^3})+10(z-\frac{1}{z})$$ and yes you have right!

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