if a + b + c = 0 then prove that a cube plus b cube plus c cube equal to 3 ABC
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Heya,
Given, a + b + c = 0
=>a + b = -c ---------------- (1)
cubing both sides, we get
(a+b)^3 = (-c)^3
a^3 + b^3 + 3ab(a+b) = -c^3
a^3 + b^3 + c^3 = -3ab(a+b)
a^3 + b^3 + c^3 = -3ab(-c) (from (1))
a^3 + b^3 + c^3 = 3abc.
Hence, Proved!
HOPE IT HELPS:-))
Given, a + b + c = 0
=>a + b = -c ---------------- (1)
cubing both sides, we get
(a+b)^3 = (-c)^3
a^3 + b^3 + 3ab(a+b) = -c^3
a^3 + b^3 + c^3 = -3ab(a+b)
a^3 + b^3 + c^3 = -3ab(-c) (from (1))
a^3 + b^3 + c^3 = 3abc.
Hence, Proved!
HOPE IT HELPS:-))
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