i need this questions answers now pls Share with your friends Share 0 Manbar Singh answered this Dear Student, Let ∆ = a2bcac+c2a2+abb2acabb2+bcc2Taking a,b,c common from C1, C2 and C3 respectively∆ = abcaca+ca+bbabb+ccApplying C1 = C1 + C2 + C3, we get ∆ = abc2a+cca+c2a+bba2b+cb+cc⇒∆ = 2abca+cca+ca+bbab+cb+ccApply C2 = C2 - C1 and C3 = C3 - C1⇒∆ = 2abca+c-a0a+b-a-bb+c0-bApplying C1 = C1 + C2 + C3, we get ∆ = 2abcc-a00-a-bc0-b⇒∆ = 2abc × abc1-100-1-110-1⇒∆ = 2a2b2c211-0 + 10+1⇒∆ = 4a2b2c2 Regards 7 View Full Answer
