Question

plz solve this ques

fast fast fast fast​​

Answer 1

Answer:

heyy

b^ 2+c^2-a^2

=>(b+c)^2–2bc-a^2 [(b+c)^2=b^2+c^2+2bc]

=>(b+c+a)(b+c-a)-2bc [a^2-b^2=(a+b)(a-b)]

=>b^2+c^2-a^2 = -2bc [ a+b+c=0]

Similarly, we can prove

a^2+b^2-c^2=-2ab

a^2+c^2-b^2=-2ac

Putting the above values in given equation

=> 1/2(-1/bc-1/ca-1/ab)

=> -1(a+b+c)/2abc [a+b+c=0]

=> 0

please mark brainliest if this answer is correct!