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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