J595 Mathematical Reflection
\sqrt[3]{(x-1)^2} - \sqrt[3]{2(x-5)^2} + \sqrt[3]{(x-7)^2} = \sqrt[3]{4x}
By Titu Adreescu.
My Solution (potentially wrong):
Let A = \sqrt[3]{(x-1)^2} , B = \sqrt[3]{2(x-5)^2}, C = \sqrt[3]{(x-7)^2} , D = \sqrt[3]{4x}.
Hence we have A + C = B + D.
Notice that A^3 + C^3 = B^3 + D^3 .
Also notice that (A+C)^3 = (B+D)^3 yields AC = BD .
Also by factoring out A^3 + C^3 = (A+C)(A^2-AC-C^2) and doing the same for B^3+D^3 gives us A^2 + C^2 = B^2 + D^2.
Lemma 1: (A^3 - C^3)^2 = (B^3 - D^3)^2.
Proof: Note that (A^3+C^3)^2 - (A^3-C^3)^2 = 4A^3C^3 = 4B^3D^3 = (B^3+D^3)^2 - (B^3-D^3)^2 and the conclusion follows. \square
Using Lemma 1 we get ( (x-1)^2 - (x-7)^2)^2 = (2(x-5)^2 - 4x)^2 and solving for x gives us x = (9 \pm 4\sqrt{2}) , (3 \pm 2\sqrt{2}) . \square
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