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Algebra Difficulty 4.7 AIME Prove it Brazil

Let aa, bb and cc be real numbers such that a≠ba \neq b and a2(b+c)=b2(c+a)=2010a^2(b + c) = b^2(c + a) = 2010. Compute c2(a+b)c^2(a + b).

Solution

Since a≠ba \neq b, a2(b+c)=b2(c+a)  ⟺  a2b+a2c−b2c−ab2=0  ⟺  ab(a−b)+c(a−b)(a+b)=0  ⟺  ab+ca+bc=0a^2(b+c) = b^2(c+a) \iff a^2b + a^2c - b^2c - ab^2 = 0 \iff ab(a-b) + c(a-b)(a+b) = 0 \iff ab+ca+bc = 0.

So (a−c)(ab+bc+ca)=0  ⟺  a2b+abc+a2c=abc+bc2+ac2  ⟺  c2(a+b)=a2(b+c)=2010(a-c)(ab+bc+ca) = 0 \iff a^2b + abc + a^2c = abc + bc^2 + ac^2 \iff c^2(a+b) = a^2(b+c) = 2010.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.