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 $x = (9 \pm 4\sqrt{2}) , (3 \pm 2\sqrt{2})$ . $\square$

Maths by Chance