Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Philippines

Problem:
Find all triples of positive real numbers (x,y,z)(x, y, z) which satisfy the system
{x3−y3−z3=64x4−y4−z4=32x6−y6−z6=8 \begin{cases} \sqrt[3]{x} - \sqrt[3]{y} - \sqrt[3]{z} = 64 \\ \sqrt[4]{x} - \sqrt[4]{y} - \sqrt[4]{z} = 32 \\ \sqrt[6]{x} - \sqrt[6]{y} - \sqrt[6]{z} = 8 \end{cases}

Solution

Solution:
Using the first and the third equations, we find that
(8+y6+z6)2=64+y3+z3 (8 + \sqrt[6]{y} + \sqrt[6]{z})^2 = 64 + \sqrt[3]{y} + \sqrt[3]{z}
Simplifying, we obtain the following equation
8y6+8z6+yz6=0 8 \sqrt[6]{y} + 8 \sqrt[6]{z} + \sqrt[6]{y z} = 0
which has no solution.

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