Question #56893

1 Answer
Mar 18, 2017

((a,b,c),(1,sqrt(2)/2,sqrt(2)/2),(1,-sqrt(2)/2,-sqrt(2)/2),(-1,sqrt(2)/2,-sqrt(2)/2),(-1,-sqrt(2)/2,sqrt(2)/2))

Explanation:

Making a=A,b=B/sqrt(2), c=C/sqrt(2) and substituting we have

A^4+B^4+C^4-4ABC=-1

Making A=B=C=X (because of symmetry) we have

3X^4-4X^3+1=0

by inspection we have the solutions

X=1, X=1,X=1/3(-1pmi sqrt(2)) so the real solutions are

A=B=C=1 or

a=1,b=sqrt(2)/2,c=sqrt(2)/2

There are four symmetries so there are four solutions

((a,b,c),(1,sqrt(2)/2,sqrt(2)/2),(1,-sqrt(2)/2,-sqrt(2)/2),(-1,sqrt(2)/2,-sqrt(2)/2),(-1,-sqrt(2)/2,sqrt(2)/2))