if x^3=1,then det.=|a b c| equals?|b c a||c a b|

You haven''t mentioned what is x = ? how ''x'' is related to a,b,c ?
det of  
a    b    c
b    a    c
c    a    b
is 3abc-a^3-b^3-c^3
and if a,b,c are roots of x^3=1 then value of this det.=0
since a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
for x^3=1 , x=1 , (-1+i√3)/2, (-1-i√3)/2 or 1, ω, ω2
so a+b+c= 1+ω+ω2=0
=> det. =0