if a+b+c=0 then find value of 1/a^2-bc+1/b^2-ca+1/c^2-ab
Answers
Answered by
1
If a+b+c=0 then av2-bc, bv2-ac, cv2-ab are, each greater than o.
Hence 1/(av2-bc)+1 /(bv2-ac)+1 /(cv2-ab)=
3/(av2-bc)+(bv2-ac) +(cv2-ac)
(1/x+1/y+1/z)=3 /(x+y+z)
Hence 1/(av2-bc)+1 /(bv2-ac)+1 /(cv2-ab)=
3/(av2-bc)+(bv2-ac) +(cv2-ac)
(1/x+1/y+1/z)=3 /(x+y+z)
Similar questions