If x + y + z = 0, prove that|xa yb zc| |a b c||yc za xb|= xyz |c a b||zb xc ya| |b c a|

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Your question is not correct it should be, if x+y+z=0 then, xaybzcyczaxbzbxcya=xyzabccabbcaNow by expansion, L.H.S.= xayza2-x2bc-yby2ac-xzb2+zcxyc2-z2ab=xyza3+b3+c3-abcx3+y3+z3.....(1)Now x+y+z=0so x3+y3+z3=3xyz putting in (1) we get, Δ=xyza3+b3+c3-3abcNow taking determinant of R.H.S, we get=xyza(a2-bc)-b(ca-b2)+c(c2-ab)=xyza3+b3+c3-3abc=L.H.S

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