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I have $\frac{2ab}{ab-ac-bc+c^2}$. Of course it's $\frac{2ab}{c^2-ac-bc+ab}$ which is $\frac{2ab}{c^2-c(a+b)+ab}$ but for the latter Wolfram tells me it's not the same as $\frac{2ab}{ab-ac-bc+c^2}$. Why? What's wrong with it?

And I know it probably sounds as newbish as it can but I don't have the slightest idea what's wrong here.

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  • $\begingroup$ What precisely was your input to Wolfram Alpha? Do you have a screenshot of what you saw? $\endgroup$ – J. M. is a poor mathematician May 7 '12 at 13:50
  • $\begingroup$ My input was ((ab-ac-bc+c^2)-(c^2-c(a+b)+ab)). While for ((ab-ac-bc+c^2)-(c^2-ca-cb+ab)) it outputs zero, for my output it does not. $\endgroup$ – Straightfw May 7 '12 at 13:53
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    $\begingroup$ Likely it's just the thing being careful, since $a,b,c$ can be zero, for instance... $\endgroup$ – J. M. is a poor mathematician May 7 '12 at 13:56
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    $\begingroup$ @JM: The first input Sfw states s/he used does not involve division like the actual expressions in the question, though a couple of W|A's responses are hilariously dumb. $\endgroup$ – anon May 7 '12 at 14:00
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The problem is that $c(a+b)$ is being interpreted as a function application. If you put a space after $c$, it works. That underlines the point that you should have included your precise input in the original question.

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  • $\begingroup$ In Mathematica, the ambiguity between implied multiplication with parentheses and function application is deftly sidestepped, with brackets being restricted to function application, and parentheses for grouping expressions. I suppose Alpha is poor at talking about its assumptions with the input given... $\endgroup$ – J. M. is a poor mathematician May 7 '12 at 14:33

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