Divide by x³ and y³−5y = 0 — Six Solutions Hide in One Move | Math Olympiad

Reciprocal substitution, cubic factoring, olympiad algebra — a sextic equation with mixed positive and negative powers reveals a clean cubic the moment you divide by x³ and introduce y = x+1/x. (x⁶+1)/x² = 2x²+2. Multiply by x² and rearrange to get x⁶−2x⁴−2x²+1 = 0. This looks like a sixth-degree polynomial with no obvious factorization — until you divide by x³. Regroup the terms and you get (x³+1/x³)−2(x+1/x) = 0. Set y = x+1/x and use the identity x³+1/x³ = y³−3y. Everything collapses to y³−5y = 0, which factors as y(y²−5) = 0. Three values of y: y = 0 gives x²+1 = 0, producing two pure imaginary roots x = ±i. y = √5 gives x²−√5·x+1 = 0 by the quadratic formula, yielding x = (√5±1)/2 — two real irrational roots including the golden ratio φ = (√5+1)/2. y = −√5 gives x²+√5·x+1 = 0, producing two more real roots: x = (1−√5)/2 and x = −(1+√5)/2. Six solutions total: two pure imaginary, four real irrational. The perfect cubic y³−5y = 0 is what makes the problem tractable — without the substitution y = x+1/x, the sixth-degree polynomial has no visible entry point. The division by x³ is the key insight. Most students see a sextic and expect expansion or numerical methods. The reciprocal structure and the y substitution expose a cubic that factors completely, delivering all six roots analytically. This reciprocal-division technique — divide by the middle power, introduce y = x±1/x, exploit the identity for sums of reciprocal powers — applies to any even-degree polynomial where reciprocal terms balance. The degree always halves under this substitution. 🔔 Daily olympiad problems — algebra, number theory, geometry, combinatorics. #OlympiadMath #ReciprocalSubstitution #SexticEquation #CubicFactoring #ComplexRoots #AlgebraTrick #CompetitionMath #AMCPrep #AIМЕPrep #MathOlympiad #GoldenRatio #OlympiadAlgebra