# a/(a-b)(a-c) + b/(b-c)(b-a) + c/(c-a)(c-b)

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### 2 Answers

We simplify the expression involving the three terms each being a rational algebraic expression. We also assume that the given expression is a/((a-b)(a-c)) +b/((b-c)(b-a))+c/((c-a)(c-b))

Then each ter can have an equivalent expression for each of term by taking the common dinominator as the LCM of the dinominators of the 3 terms, i.e. (a-b)(b-c)(c-a).

Then the 1st term:a/((a-b)(a-c)) = a(b-c)(-1)/{(a-b)(b-c)(c-a)}

2nd term: b/((b-c)(b-a))= b(-1)(c-a)/{(a-b)(b-c)(c-a)}

3rd term: c/((c-a)(c-b))= c(-1)(a-b)/{(a-b)(b-c)(c-a)]

Therfore, the dinominators being same we can add the numerators of the equivalent expressions.

Su of the numerators; -a(b-c)-b(c-a)-c(a-b)= -ab+ac-bc+ab-ca+bc= 0

Therefore, the sum of the three terms = 0/{(a-b)(b-c)(c-a)}=0.

a/(a-b)(a-c) + b/(b-c)(b-a) + c/(c-a)(c-b)

= a/(a-b)(a-c) + b/[-(b-c)(a-b)] + c/(a-c)(b-c)

= [a(b-c) - b(a-c) + c(a-b)]/[(a-b)(a-c)(b-c)]

= (ab -ac -ab +bc +ac -bc)/[(a-b)(a-c)(b-c)]

= 0/[(a-b)(a-c)(b-c)] = 0