If a,b,c are all non-zero and a+b+c=0, prove that a2bc+b2ac+c2ab=3
Given,
a+b+c=0
or, a+b=−c
cubing both sides we get,
(a+b)3=−c3
a3+b3+3ab(a+b)=−c3 [∵(a+b)3=a3+3a2b+3ab2+b3]
a3+b3−3abc=−c3 [Since a+b=−c]
a3+b3+c3=3abc.......(1).
Now,
a2bc+b2ac+c2ab=a3+b3+c3abc
=3abcabc [Using (1)]
=3.