Question

If a+b+c =0 show that a³+b³+c³= 3abc

Answer 1

Answer:

Step-by-step explanation:

a+b+c=0, a+b=-c,a+c=-b, b+c=-a

(a+b+c)^3=(a+b+c)^2*(a+b+c)=(a^2+b^2+c^2+2ab+2ac+2bc)*(a+b+c)=

a^3+ab^2+ac^2+2a^2b+2a^2c+2abc+a^2b+b^3+bc^2+2ab^2+2abc+2b^2c+a^2c+b^2c+c^3+2abc+2ac^2+2bc^2=a^3+b^3+c^3+3a^2b+3a^2c+3ac^2+3ab^2+3bc^2+3b^2c+6abc=

a^3+b^3+c^3+3a^2*(b+c)+3c^2(a+b)+3b^2(a+c)+6abc=

a^3+b^3+c^3+3a^2*(-a)+3c^2*(-c)+3b^2*(-b)+6abc=

a^3+b^3+c^3-3a^3-3c^3-3b^3+6abc=

6abc-2a^3-2b^3-2c^3=2(3abc-a^3-b^3-c^3)=

2*[3abc-(a^3+b^3+c^3)]=0

so 3abc-(a^3+b^3+c^3)=0

so a^3+b^3+c^3=3abc