If a, b and c are all non-zero and a + b + c = 0, then find the value of a2bc+b2ac+c2ab.
3
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Solution
The correct option is A 3 We know that, a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab–bc–ca) =0(a2+b2+c2–ab–bc–ca)[∵a+b+c=0,given] =0 →a3+b3+c3=3abc
On dividing both sides by abc; we get, a3abc+b3abc+c3abc=3 ⇒a2bc+b2ac+c2ab=3