**Problem**: (Math Olympiad Treasures - 1.1.10) Find the locus of points $ (x,y) $ for which

$ x^3+y^3+3xy = 1 $.

**Solution**: Recall the useful factorization

$ a^3+b^3+c^3-3abc = \frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2] $.

Setting $ a = x $, $ b = y $, and $ c = -1 $, we obtain

$ x^3+y^3-1+3xy = \frac{1}{2}(x+y-1)[(x-y)^2+(y+1)^2+(-1-x)^2] $.

But from the given condition this expression is zero, so either

$ x+y-1 = 0 $,

which gives us the line $ x+y = 1 $ or

$ (x-y)^2+(y+1)^2+(-1-x)^2 = 0 $,

which gives us the point $ (-1,-1) $.

So the locus of points is the line $ x+y = 1 $ and $ (-1,-1) $. QED.

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Comment: The factorization basically finished the problem; it's nice to remember them because when they apply, they are extremely useful.

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Practice Problem: (Math Olympiad Treasures - 1.1.6) Let $ a,b,c $ be distinct real numbers. Prove that the following equality cannot hold:

$ \sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a} = 0 $.

Set x = cbrt(a-b), y = cbrt(b-c), z = cbrt(c-a). Then x^3 + y^3 + z^3 = 0. Moreover, from the factorization we have

ReplyDelete-3xyz = 1/2(x + y + z)( (x-y)^2 + (y-z)^2 + (z-x)^2 ). The LHS cannot be zero because a, b, c are distinct, hence x + y + z \neq 0. QED.

How do you know all these factorizations? I've never even heard of that one.

ReplyDeleteWell I learned this one from the book that I got the problem from, but it's shown up a few times (there was a USAMTS question related to it last year?).

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