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Algebra

  • Titu AndreescuEmail author
  • Bogdan Enescu
Chapter
  • 2.5k Downloads

Abstract

A very useful algebraic identity is derived by considering the following problem.

Problem 1.1 Factor a 3+b 3+c 3−3abc.

Solution Let P denote the polynomial with roots a,b,c:
$$P(X)=X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc.$$
Because a,b, c satisfy the equation P(x)=0, we obtain
$$\begin{array}{*{20}{c}} {{a^3} - \left( {a + b + c} \right){a^2} + \left( {ab + bc + ca} \right)a - abc = 0,} \\ {{b^3} - \left( {a + b + c} \right){b^2} + \left( {ab + bc + ca} \right)b - abc = 0,} \\ {{c^3} - \left( {a + b + c} \right){c^2} + \left( {ab + bc + ca} \right)c - abc = 0.} \end{array}$$
Adding up these three equalities yields
$$a^{3}+b^{3}+c^{3}- (a+b+c )\bigl(a^{2}+b^{2}+c^{2}\bigr)+(ab+bc+ca)(a+b+c)-3abc=0.$$
Hence Note that the above identity leads to the following result: if a+b+c=0, then a 3+b 3+c 3=3abc.

Keywords

Algebraic Identities Rearrangement Inequality Well-known Cauchy-Schwarz Inequality Proposed Problem Nonzero Real Number 
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

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