Question

when a+b+c=0, then how a^3+b^3+c^3-3abc=0

Answer 1

Sol : Given ,

 a + b + c = 0 

 a + b = - c     

By cubing both sides,

 ( a + b )³ = ( - c )³

 a³ + b³ + 3 ab ( a + b ) = - c³

By substituting the value of ( a + b ) in equation 1,

 a³ + b³ + 3 ab ( - c ) = - c³

 a³ + b³ + c³ - 3 abc = 0.Proved.

  
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Answer 2

hiii friend
here is your answer
as, a + b + c = 0
then, a + b = -c
and now, by cubing both sides, we get;
(a + b)^3 = (-c)^3
${a}^{3} + {b}^{3} + 3ab(a + b) = {( - c)}^{3}$
as, a + b = - c

${a}^{3} + {b}^{3} + 3ab( - c) = {( - c)}^{3} \\ \\ {a}^{3} + {b}^{3} - 3abc= {( - c)}^{3} \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc= 0 \\ \\ hence \: \: proved$


glad to help you
hope it helps
thank you.