Question 250665
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I presume that you want the factorization of *[tex \LARGE (x\,-\,y)^3\ -\ y^3]

This is clearly the difference of two cubes as you said, so let's look at the pattern:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\,\pm\,b^3\ =\ \left(a\,\pm\,b\right)\left(a^2\,\mp\,ab\,+\,b^2\right)]

Now, let *[tex \Large x\,-\,y\ =\ a] and *[tex \Large y\ =\ b] and make all of the appropriate substitutions, including sign selections:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\,-\,y)^3\ -\ y^3\ =\ \left((x\,-\,y)\,-\,y\right)\left((x\,- \,y)^2\,+\,(x\,-\,y)y\,+\,y^2\right)]

Then we do a little simplification:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,2y\right)\left(\left(x^2\,-\,2xy\,+\,y^2\right)\,+\,xy\,-\,y^2\,+\,y^2\right)]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,2y\right)\left(x^2\,-\,xy\,+\,y^2\right)]

And that, as they say, is that.  By the way, the mnemonic for remembering the signs in the sum and difference of two squares factorization is San Diego Padres - SDP - Same, Different, Positive.  The sign in the binomial term is the same as the the sign in the thing you are trying to factor, the first sign in the trinomial is the opposite sign, and the last sign is always positive.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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