Question

show that if a+b+c=0 then a3+b3+c3=3 abc

Answer 1

a+b+c =0 (given)

a+b = -c -------(i)

cube both sides


(a+b)^3 = -c^3
a^3 + b^3 + 3ab(a+b) = -c^3
usibg eq. 1

a^3 + b^3 +3ab(-c)= -c^3
a^3 +b^3 -3abc = -c^3
exchange sides and you'll get

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