prove that a³+b³+8c³=6abc?
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Answer:
proved.
Step-by-step explanation:
Given: a + b + 2c =0
by taking whole cube
(a + b + 2c)³ = 0
=> (a+b)³ + 3(a+b)²2c + 3(a+b)(2c)² + (2c)³ = 0
=> a³ + 3a²b + 3ab² + b³ + 3(2a²c + 2b²c + 4ac² + 4bc²) + 12abc + 8c³ = 0
=> a³ + b³ + 8c³ + 3a²(b+2c) + 3b²(a+2c) + 3(2c)²(a+b) + 12abc = 0
=> a³ + b³ + 8c³ + 3(a+b)(b+2c)(2c+a) = 0
=> a³ + b³ + 8c³ + (a + b + 2c) [ a² + b² + 4c² -ab -2bc - 2ca] - 3ab(2c) = 0
since (a+b+2c) = 0
therefore,
a³+b³+8c³ - 3ab(2c) = 0
=> a³ + b³ + 8c³ = 6abc
Hence, proved
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