Question

If a,b and c are all non-zeros real no.s and a+b+c =0 .
Then prove that ,
(a²/bc)+(b²/ca)+(c²/ac)=3

Answer 1

(a²/bc)+(b²/ca)+(c²/ac)
=(a^3+b^3+c^3)/abc [By LCM and addition of fractions]
Since a+b+c=0,
Therefore a^3 +b^3+c^3=3abc
(a^3+b^3+c^3)/abc=3abc/abc
=3
Since LHS=RHS,
Hence proven.
Plz mark as brainliest :D

Answer 2

(a²/bc)+(b²/ca)+(c²/ac)=3

LHS

(a²/bc)+(b²/ca)+(c²/ac)

(a×a²+b× b²+c× c²)/abc

(a³+ b³+c³)/abc


[If a+b+c= 0 then (a³+ b³+c³)= 3abc]

3abc/abc

= 3

LHS= RHS