if a+b+c=0
then a²/bc+b²/ca+c²/ab=?
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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
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