What are the conditions on real numbers *x*, *y*, *z* such that ?

Well, inspection shows us that if (*x*,*y*,*z*) is a solution, so are the other 5 permutations of that solution. We also note that if *y*=-*x*, we see that , and , so the equality holds for *y*=-*x*, and thus, via the permutations, for *z*=-*x* and *z*=-*y*. However, are there any other solutions? Some algebra allows us to check. We will use the difference of cubes formula and the sum of cubes formula . So:

.

Moving the terms to the same side,

So our equality is true if *x*+*y*=0, *x*+*z*=0, or *y*+*z*=0; precisely the solution we found by inspection.

### Like this:

Like Loading...

*Related*

Tags: Math, Monday Math

This entry was posted on May 18, 2009 at 1:35 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

## Leave a Reply