解： (a+b+c)³=Σ(i,j,k=0,1,2,3且i+j+k=3)[3!/(i!j!k!)]·a^i·b^j·c^k=a³+b³+c³+3a²b+3ab²+3a²c+3ac²+3b²c+3bc²+6abc.
(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc
