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# Algebra

• Titu Andreescu
• Bogdan Enescu
Chapter
• 2.4k Downloads

## Abstract

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
$$\begin{gathered} {\left( {a + 2b - 3c} \right)^3} + {\left( {b + 2c - 3a} \right)^3} + {\left( {c + 2a - 3b} \right)^3} \hfill \\ \quad = 3\left( {a + 2b - 3c} \right)\left( {b + 2c - 3a} \right)\left( {c + 2a - 3b} \right). \hfill \\ \end{gathered}$$

## Keywords

Rearrangement Inequality Algebraic Identities Convex Function Algebraic Geometry Absolute Value
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

## Copyright information

© Springer Science+Business Media, LLC 2011

## Authors and Affiliations

1. 1.School of Natural Sciences and MathematicsUniversity of Texas at DallasRichardsonUSA
2. 2.Department of Mathematics“BP Hasdeu” National CollegeBuzauRomania