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Wolfram Alpha tells me that $$\frac{a+1}{1-a^2}=\frac{1}{1-a}.$$ but can't see how! Any help would be very appreciated!

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Multiply the equation throughout by the factor $1-a$, and the result follows, when 1-a is not zero. – awllower Nov 6 '12 at 17:58
up vote 2 down vote accepted

Hint: Factor $1-a^2$ using the difference of squares formula: $(a-b)(a+b)=a^2-b^2$.

It is also worth noting that, strictly speaking, that equation is not true for all $a$, but it is when $a\neq -1$ and $a \neq 1$.

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@m.k. It's true that $a\neq 1$ eliminates a case where both sides are undefined, but in almost all texts an equation is considered true if it holds whenever both sides are defined. The point was that $a+1$ can't be cancelled if $a=-1$. That is different from the $a=1$ case. – rschwieb Nov 6 '12 at 17:43

$$ \frac{a+1}{1-a^2} = \frac{a+1}{(1+a)(1-a)} = \frac{1}{1-a} $$

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Perhaps you could avail of the teX format? For example, try $(a+1)/(1-a^2 ) = (a+1)/((1+a)(1-a)) = 1/(1-a)$ – awllower Nov 6 '12 at 18:00

Use $(x+y)(x-y) = x^2 - y^2$ on $1 - a^2$.

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$$\frac{a+1}{1-a^2}=\frac{a+1}{1-a+a-a^2}=\frac{a+1}{1(1-a)+a(1-a)}=$$ $$=\frac{a+1}{(1-a)(1+a)}=\frac{1(1+a)}{(1-a)(1+a)}=\frac{1}{1-a}.$$

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