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