Prove that |a2 a2-(b-c)2 bc| |b2 b2-(c-a)2 ca| |c2 c2-(a-b)2 ab| = (a-b) (b-c) (c-a) (a+b+c) (a2+b2+c2) Share with your friends Share 9 Prasanta answered this To prove: a2a2-b-c2bcb2b2-c-a2cac2c2-a-b2ab=a-bb-cc-aa+b+ca2+b2+c2LHS= a2a2-b-c2bcb2b2-c-a2cac2c2-a-b2abApply R1→R1-R2 and R2→R2-R3=a-ba+b2a-ba+b-c-ca-bb-cb+c2b-cb+c-a-ab-cc2c2-a-b2abTake a-b and b-c common from R1 and R2=a-bb-ca+b2a+b-c-cb+c2b+c-a-ac2c2-a-b2abNow, apply R2→ R2- R1=a-bb-ca+b2a+b-c-cc-a4c-ac-ac2c2-a-b2abTake c-a common from R2=a-bb-cc-aa+b2a+b-c-c141c2c2-a-b2abApply C1→C1-C3=a-bb-cc-aa+b+c2a+b-c-c041c2-abc2-a-b2abNow, expand along C1, we get=a-bb-cc-a[a+b+c(4ab-c2+a-b2+c2-ab2a+b-c+4c)]=a-bb-cc-a[a+b+c(4ab-c2+a-b2)+2c2-aba+b+c]Take a+b+c common=a-bb-cc-a a+b+c[4ab-c2+a-b+2c2-2ab2] =(a-b)(b-c)(c-a)(a+b+c)(a2+b2+c2) =RHSHence proved. 6 View Full Answer
