Question #56893

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Answer 1 · 2017-03-18T15:02:48

#((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))#

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))#