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3s=a+b+c if (s-a)²/(s-b)(s-c)+(s-b)²/(s-c)(s-a)+(s-c)²/(s-a)(s-b)=?

3s=a+b+c if (s-a)²/(s-b)(s-c)+(s-b)²/(s-c)(s-a)+(s-c)²/(s-a)(s-b)=?

Grade:12

1 Answers

Shubhang Kulkarni
33 Points
one year ago
By taking LCM to equalise denominator: 
[(s-a)3 + (s-b)3 + (s-c)3] / [(s-a)(s-b)(s-c)] 
 
We know that, x3+y3+z3 = 3xyz when x+y+z = 0 
In above expression: 
(s-a)+(s-b)+(s-c) = 3s-a-b-c
(s-a)+(s-b)+(s-c) = a+b+c-a-b-c   (Since 3s = a+b+c)
(s-a)+(s-b)+(s-c) = 0
 
Therefore, the expression can be simplified to: 
3[(s-a)+(s-b)+(s-c)] / [(s-a)+(s-b)+(s-c)]
= 3
 
Thanks, hope the solution helps. 
 

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