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Is there a name for this "trick" or more general method to know that the product of the numerators is that cubic difference? Other than simply multiplying through and cancelling.
I'm pretty sure someone explained to me a way to know it.
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>>8463308
Use (a + b)^n =(n choose k)a^kb^(n-k)
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>>8463311
Not sure I see it. Wouldn't it need to be 2ab for binomial theorem?
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>>8463308
I'd stick with multiplying
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>>8463308
its called polynomial long division
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[eqn]x^{n+1}-y^{n+1}=(x-y)\sum_{i=0}^n x^i y^{n-i}[/eqn]
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