If a+b+c = 0, P.T. a²/ bc + b²/ca +c²/ab =3

a+b+ c=0

=>a³ +b³ +c³ = 3abc......(i)

a²/bc = 3a³/3abc ...........(ii)

b²/ca =3b³/3abc . . . . . .(iii)

c²/ab =3c³/abc . . . . . . . (iv)

=>from (i), (ii), (iii) & (iv),

a²/bc +b²/ca +c²/ab = 3(a³+b³+c³) /3abc

=>3. . . . . . . . . . . . . . . . . . (Ans)

thumbzz up plzz :)

thanx debasmita !!!!!

one sec hw can the answer be 3?

maybe the answer is a²+b²+c²

@flower,there are common number [second,third and fourth steps] just do and see...