Question

cho a+b+c=0. Chứng minh a^3+b^3+c^3=3abc

Answer 1

Ta có :(a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²c+3b²a+3c²a+3c²b+6abc

            (a+b+c)³=a³+b³+c³+(3a²b+3a²b+3abc)+(3b²c+3b²a+3abc)+(3c²a+3c²b+3abc)-3abc

            (a+b+c)³=a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)-3abc

            (a+b+c)³=a³+b³+c³+3(a+b+c)(ab+bc+ac)-3abc

  thay a+b+c=0 ta được 

              0³=a³+b³+c³+3.0(ab+bc+ac)-3abc

             0=a³+b³+c³-3abc

=>a³+b³+c³=3abc

Answer 2

a+b+c=0

=>a+b=-c

=>(a+b)³=-c³

=>a³+b³+3a²b+3ab²+c³=0

=>a³+b³+c³+3ab(a+b)=0

Mà a+b=-c

=>a³+b³+c³+3ab.(-c)=0

=>a³+b³+c³-3abc=0

=>a³+b³+c³=3abc