-Demonstre que:
a³ + b³ + c³ − 3abc =
(a + b + c)(a² + b² + c² − ab − bc − ac).
a³ + b³ + c³ − 3abc =
(a + b + c)(a² + b² + c² − ab − bc − ac).
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Vamos là.
(a + b + c)(a² + b² + c² - ab − bc - ac) =
a³ + ab² + ac² - a²b - abc - a²c + ba² + b³ + bc² - ab² - b²c - abc + ca² + cb² + c³ - abc - bc² - ac²
cancelar os quadrados fica
a³ + b³ + c³ - 3abc