Question

a+b+c=0 then the value of a²/bc+b²/ac+c²/ab​

Answer 1

Answer:

a2bc+b2ac+c2ab=a3+b3+c3abc—(1)

a3+b3+c3=(a+b+c)3−3ab(a+b)−3bc(b+c)−3ac(a+c)−6abc—(2)

Since a+b+c=0,b+c=−a,c+a=−b,a+b=−c—(3)

Substitute (3)in(2),

=>a3+b3+c3=03−3ab(−c)−3b(−a)−3ac(−b)−6abc

=>3abc+3abc+3abc−6abc=3abc—(4)

From(1)and (4),a3+b3+c3abc=3abcabc=3

Ans: 3