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I want to know how to expand the $$\prod_{j=1}^{r}x_{j}^{-1}=\sum_{j=1}^{r}x^{-1} \prod_{j=1,k \neq j}^{r}x_{k}^{-1}$$ where $x=\sum_{j=1}^{r}x_{j}$.

I tried to do this for $r=3$ $$\frac{1}{x_1x_2x_3}=\frac{1}{x_1} \frac{1}{x_2x_3}+\frac{1}{x_2} \frac{1}{x_1x_3}+\frac{1}{x_3} \frac{1}{x_1x_2}=\frac{3}{x_1x_2x_3}$$ but there is something wrong.

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I tried to do this for $r=3$ $$\frac{1}{x_1x_2x_3}=\frac{1}{x_1} \frac{1}{x_2x_3}+\frac{1}{x_2} \frac{1}{x_1x_3}+\frac{1}{x_3} \frac{1}{x_1x_2}$$ but there is something wrong.

Your misread $x$ as $x_i$, so the right expansion is

$$\frac{1}{x_1x_2x_3}=\frac{1}{x_1+x_2+x_3} \frac{1}{x_2x_3}+\frac{1}{x_1+x_2+x_3} \frac{1}{x_1x_3}+\frac{1}{x_1+x_2+x_3} \frac{1}{x_1x_2}$$

After multiplying both parts by $x_1\dots x_3$ we obtain

$$1=\frac{x_1}{x_1+x_2+x_3}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_1+x_2+x_3}=\frac{x_1+x_2+x_3}{x_1+x_2+x_3}=1.$$

The general case can be proved similarly:

$$\prod_{j=1}^{r}x_{j}^{-1}=\sum_{j=1}^{r}x^{-1} \prod_{j=1,k \neq j}^{r}x_{k}^{-1}$$

After multiplying both parts by $x_1\dots x_n$ we obtain

$$1=\sum_{j=1}^{r}x^{-1} x_{j}=x^{-1}\sum_{j=1}^{r} x_{j}=x^{-1}x=1.$$

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    $\begingroup$ I understood. Thank you very much! $\endgroup$ – Oily Nov 12 '16 at 13:49
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    $\begingroup$ I was confused by the fact that $k \neq j$ $\endgroup$ – Oily Nov 12 '16 at 13:56

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