a |
b−c |
b |
c−a |
c |
a−b |
b+c |
(c−a)+(a−b) |
b+c |
c−b |
a |
b−c |
b+c |
c−b |
b+c |
b−c |
∴a=-(b+c),
∴a3+b3+c3
=[-(b+c)]3+b3+c3
=-(b+c)3+(b+c)(b2-bc+c2)
=(b+c[-(b+c)2+b2-bc+c2]
=(b+c)(-b2-2bc-c2+b2-bc+c2)
=-a•(-3bc)
=3abc.
∴a3+b3+c3=3abc(证毕).
a |
b−c |
b |
c−a |
c |
a−b |
a |
b−c |
b |
c−a |
c |
a−b |
b+c |
(c−a)+(a−b) |
b+c |
c−b |
a |
b−c |
b+c |
c−b |
b+c |
b−c |