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I'm currently working on a physics problem and I have worked out the problem to the reach the following expression

$$\dfrac{2\epsilon(e^{-\epsilon\beta}+e^{-2\epsilon \beta})}{1+2e^{-\epsilon \beta}+ e^{-2 \epsilon \beta}}$$

Is there a way that I can turn the expression that I have into this

$$\dfrac{2\epsilon}{1+e^{\epsilon \beta}}$$

This is the answer in the book and I don't have a clue how it got there to be honest.

EDIT: Thanks very much guys!!

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  • $\begingroup$ Mathematica simplifies the expression immediately. $\endgroup$ – David G. Stork Oct 5 '17 at 21:14
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$$ \frac{2\epsilon(e^{-\epsilon\beta}+e^{-2\epsilon \beta})}{1+2e^{-\epsilon \beta}+ e^{-2 \epsilon \beta}} $$

$$ = \left(\frac{2\epsilon(e^{-\epsilon\beta}+e^{-2\epsilon \beta})}{1+2e^{-\epsilon \beta}+ e^{-2 \epsilon \beta}} \right) \frac{e^{2 \epsilon \beta}}{e^{2 \epsilon \beta}} $$

$$ = \frac{2\epsilon(e^{\epsilon\beta}+1)}{e^{2 \epsilon \beta}+2e^{\epsilon \beta}+ 1} $$

$$ = \frac{2\epsilon(e^{\epsilon\beta}+1)}{(e^{\epsilon \beta}+1)^2} $$

$$ = \frac{2\epsilon}{(1+e^{\epsilon \beta})} $$

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$$\dfrac{2\epsilon(e^{-\epsilon\beta}+e^{-2\epsilon \beta})}{1+2e^{-\epsilon \beta}+ e^{-2 \epsilon \beta}} =\dfrac{2\epsilon(e^{-\epsilon\beta}+e^{-2\epsilon \beta})}{1+2e^{-\epsilon \beta}+ e^{-2 \epsilon \beta}}\times \frac{ e^{2 \epsilon \beta}}{e^{2 \epsilon \beta}} =\dfrac{2\epsilon(e^{\epsilon\beta}+1)}{1+2e^{\epsilon \beta}+ e^{2 \epsilon \beta}} =\dfrac{2\epsilon(e^{\epsilon\beta}+1)}{(1+e^{\epsilon \beta})^2} =\dfrac{2\epsilon}{1+e^{\epsilon \beta}} $$

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HINT: Multiply numerator and denominator both by $e^{2\epsilon \beta}$ and try to simplify.

Hope this helps you.

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