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05 Oct 2019, 08:43
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\(\frac{x^3-1}{1-x^5}\)
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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05 Oct 2019, 10:27
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\(x^5 - 1 = (x-1)(x^4+x^3+x^2+x+1)\)
\(x^3 - 1 = (x-1)(x^2+x+1)\)
so \(\frac{x^3-1}{1-x^5}\) = \(\frac{(x^2+x+1)}{-(x^4+x^3+x^2+x+1)}\)
This is the simplest form, however this won't be needed in GMAT.
What we need to know is that if n is odd, then \(x^n - y^n\) is divisible by \(x-y\), and \(x^n + y^n\) is divisible by \(x+y\)
\(x^3 - 1 = (x-1)(x^2+x+1)\)
so \(\frac{x^3-1}{1-x^5}\) = \(\frac{(x^2+x+1)}{-(x^4+x^3+x^2+x+1)}\)
This is the simplest form, however this won't be needed in GMAT.
What we need to know is that if n is odd, then \(x^n - y^n\) is divisible by \(x-y\), and \(x^n + y^n\) is divisible by \(x+y\)
05 Oct 2019, 17:09
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AlvinReyes995
Is this a GMAT question??
It doesn’t seem like.
Can you pls share the source of this question
Posted from my mobile device
