Question

If a+B+c=0,
show that : a³+b³+c³=3abc.

Help.

Answer 1

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COOL QUESTION.
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Solution:-

Given :

a+b+c=0

=> a+b=-c

=> (a+b)³ = (-c)³

=> a³+b³+3ab(a+b)= (-c)³

=> a³+b³ +3ab (-c)=-c³

{Since, a+ b = -c}

Therefore,

=> a³+b³-3ab=-c³

=> a³+b³+c³=3ab.

THANKS!!

Answer 2

               

$a+b+c=0$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac) \\ \\ \Rightarrow\ a^3+b^3+c^3-3abc=0(a^2+b^2+c^2-ab-bc-ac) \\ \\ \Rightarrow\ a^3+b^3+c^3-3abc=0 \\ \\ \Rightarrow\ a^3+b^3+c^3=3abc \\ \\ \\ Hence proved!!!$

$Plz ask me if you've any doubts. \\ \\ \\ Thank you. :-)) \\ \\ \\ \\ \\ \#adithyasajeevan$