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Question
If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
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Solution
a + b + c = 0 ...(i)
⇒ (a + b) + c = 0
Cubing both sides
⇒ (a + b)3 + c3 + 3(a + b) (c) (a+ b + c) = 0
⇒ a3 + b3 + 3ab (a + b) + c3 + 0 = 0
⇒ a3 + b3 + c3 + 3ab (a + b) = 0 ...(2)
Using (i), we get, a + b = -c From (2),
a3 + b3 + c3 + 3ab (-c) = 0
⇒ a3 + b3 + c3 = 3abc.
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