If a+b+c=0 then (a2bc+b2ca+c2ab)=? (a) 1 (b) 0 (c) -1 (d) 3
a+b+c=0⇒a3+b3+c3=3abc
Thus, we have:
a2bc+b2ca+c2ab=a2×abc×a+b2×bca×b+c2×cab×c=a3abc+b3abc+c3abc=a3+b3+c3abc=3abcabc=3
(d) 3