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Problem 1.4 Factor (a+2b−3c)3+(b+2c−3a)3+(c+2a−3b)3.
Solution Observe that (a+2b−3c)+(b+2c−3a)+(c+2a−3b)=0. Because x+y+z=0 implies x 3+y 3+z 3=3xyz, we obtain
(a+2b−3c)3+(b+2c−3a)3+(c+2a−3b)3=3(a+2b−3c)(b+2c−3a)(c+2a−3b).
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