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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 cb a cc a b is 3abc-a^3-b^3-c^3and if a,b,c are roots of x^3=1 then value of this det.=0since 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, ω2so a+b+c= 1+ω+ω2=0=> det. =0
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