Problem
ALG-B1-M06-P017 Fourth powers
#17
★★★★☆ Level 4 of 5
Let \(x+y+z=0\) and \(x^2+y^2+z^2=6\). Prove that \(x^4+y^4+z^4=18\).
Find \(xy+yz+zx\), then express \(x^2y^2+y^2z^2+z^2x^2\).
From the zero sum we get \(xy+yz+zx=-3\). Then \((xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)\), and the second term is zero. Hence \(x^2y^2+y^2z^2+z^2x^2=9\).
Now \((x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)\). Therefore \(36=x^4+y^4+z^4+18\), so \(x^4+y^4+z^4=18\).
Two symmetric substitutions in a row: this is already olympiad level for the module.