Algebric identity

If a, b and c are all ≠ 0, and a + b + c = 0, then show that 

a²/_bc +b²/_ca + c²/_ba = 3

Answer:

if a+ b + c = 0, then

we have (a + b) = -c

taking cube on each side of equation, we get

a³ + b³ + 3a²b + 3ab² = -c³

or, a³ + b³ + c³ = -3ab(a + b)

or, a³ + b³ + c³ = -3ab(-c)

or, a³ + b³ + c³ = 3abc

 Dividing throughout by “abc”, we get

a²/_bc +b²/_ca + c²/_ba = 3