I am a middle school teacher and a parent. I am snowed in and trying to help my 9th grader get ready for 9 weeks exams. I have tried to factor this problem to no avail. (u-v)^3+v-u. I have the answer but I need to know how it is done. I tried writing out (u-v)(u-v)(u-v)+v-u and foiling the first two. Next I multiplied that trinomial by (u-v) that leaves me with u^3-3u^2v+3uv^2-v^3+v-u (unless I hve gone wrong somewhere). Anyway that really doesn't help me.

Any help would be appreciated.


What you need to see here is a common factor. First write (u-v)3+v-u as (u-v)3-(u-v) and then take out the common factor (u-v) to get

(u-v)3+v-u = (u-v)3-(u-v) = (u-v)[(u-v)2 - 1]

Now you need to recognize that (u-v)2-1 factors as a difference of squares

(u-v)[(u-v)2 - 1] = (u-v)[(u-v-1)(u-v+1)]
I hope this helps,
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