• Titu AndreescuEmail author
  • Bogdan Enescu


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:
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.


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