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Tuesday, September 20, 2016

Mathematical Reflections 2016, Issue 3 - Problem U374

Problem:
Let p and q be complex numbers such that two of the zeros a,b,c of the polynomial x^3+3px^2+3qx+3pq=0 are equal. Evaluate a^2b+b^2c+c^2a.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume without loss of generality that a=b. Then, a^2b+b^2c+c^2a=b^3+b^2c+bc^2. By Vi\`ete's Formulas, we have \begin{array}{rcl} abc&=&-3pq \\ ab+bc+ca&=&3q \\ a+b+c&=&-3p. \end{array} Since a=b, we have \begin{array}{rcl} b^2c&=&-3pq \\ b^2+2bc&=&3q \\ 2b+c&=&-3p. \end{array} Multiplying side by side the last two equations, we get (b^2+2bc)(2b+c)=-9pq. Since -9pq=3(-3pq)=3b^2c, we get (b^2+2bc)(2b+c)=3b^2c, i.e. b^3+b^2c+bc^2=0. It follows that a^2b+b^2c+c^2a=0.

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