Show that abc(ab+bc+ca)=a+b+c Share with your friends Share 20 Lovina Kansal answered this Dear student Let ∆ = aa3a4 - 1bb3b4 - 1cc3c4 - 1 then,∆ = aa3a4 - 1bb3b4 - 1cc3c4 - 1 + aa3-1bb3-1cc3-1⇒ ∆ = abc 1a2a31b2b31c2c3 - aa31bb31cc31 Applying R2 = R2 - R1 & R3 = R3 - R1⇒∆ = abc 1a2a31b2- a2b3 - a31c2 - a2c3 - a3 - aa31b - cb3 - a30c - ac3 - a30 ⇒∆ = abc b - a c - a 1a2a30b + ab2 + a2 + ab0c + ac2 + a2 + ac - b - a c - a aa311b2 + a2 + ab01c2 + a2 + ac0⇒ ∆ = abc b - a c - a b+aa2+b2+abc+aa2+c2+ac -b - a c - a 1a2+b2+ab1a2+c2+acApply R2 = R2 - R1⇒∆ =abc b - a c - a a+ba2+b2+abc-bc2-b2+ac-b - b - a c - a1a2+b2+ab0c2-b2+ac-b⇒∆ = abc b - a c - ac-ba+ba2+b2+ab1a+b+c - b - a c - ac-b1b - a c - a0a+b+c⇒∆ = abc b - a c - ac-ba+ba+b+c-a2+b2+ab - b - a c - ac-ba+b+c⇒∆ = abca-bb-cc-aab+bc+ca - a-bb-cc-aa+b+c⇒∆ =a-bb-cc-aabcab+bc+ca-a+b+c Now, ∆ = 0⇒a-bb-cc-aabcab+bc+ca-a+b+c = 0⇒abcab+bc+ca-a+b+c = 0 because a≠b; b≠c; c≠a, therefore a-b≠0; b-c≠0; c-a≠0⇒abcab+bc+ca = a+b+c Regards 16 View Full Answer Om Shiv answered this Experts pls solve i have given a complete question. 0