(a+b+c)³=[(a+b)+c]³=(a+b)³+3(a+b)²c+3(a+b)c²+c³
(a+b+c)³=(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c2+c³
(a+b+c)³=a³+b³+c³+3a²b+3a²c+3ab²+3b²c+3ac²+3bc²+6abc
(a+b+c)³=(a³+b³+c³)+(3a²b+3a²c+3abc)+(3ab²+3b²c+3abc)+(3ac²+3bc²+3abc)−3abc
