From \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+3ab(a+b)+b^3+c^3-3abc-3ab(a+b)=0\)

\(\Rightarrow (a+b)^3+c^3-3ab(a+b+c)=0\)

\(\Rightarrow (a+b+c)(a^2+2ab+b^2-ab-ac+c^2)-3ab(a+b+c)=0\)

\(\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\)

It's right because \(a+b+c=0\)

We have :

a³ + b³ + c³ = 3abc 

a³ + b³ + c³ - 3abc = 0

(a³ + 3a²b + 3ab² + b³) + c³ - 3a²b - 3ab² - 3abc = 0

(a + b)³ + c³ - 3ab.(a + b + c) = 0

[(a + b) + c].[(a + b)² - c.(a + b) + c²] - 3ab(a + b + c) = 0

(a + b + c).[a² + 2ab + b² - ac - bc + c²] - 3ab(a + b + c) = 0

(a + b + c).[a² + b² + c² - bc - ca + 2ab - 3ab] = 0

(a + b + c).[a² + b² + c² - ab - bc - ca] = 0

That is right , because a + b + c = 0

So a³ + b³ + c³ = 3abc

With a + b + c = 0

If \(a,b,c\in N\) then I can answer it.

\(a+b+c=0\)

=> \(a=b=c=0\).

Because \(\forall x\) in the case \(0^x=0\) and a number multiplies by \(0\) equals to \(0\).

=> \(a^3+b^3+c^3=3abc=0\left(proved\right)\)