Shubhang Kulkarni
Last Activity: 2 Years 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.