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Note that $(a,b,c)=(1,\omega,\omega^2)$ where $\omega$ is complex cube root of unity. Hence ,

$a^{1729}+b^{1729}+c^{1729}=1^{1729}+\omega^{1729}+\omega^{3458}$

Now using $\omega^3=1$ , we have:

$1^{1729}+\omega^{1729}+\omega^{3458}=1+\omega+\omega^2=\boxed{0}$

If you don't have an idea of cube roots of unity, here's an alternate approach. $x^3=1\\ ~~~~~~~~\implies (x^3)^{576}=1^{576}=1\\ \implies x^{1728}=1\\ \implies x^{1729}=x$ $\implies a^{1729}=a, b^{1729}=b, c^{1729}=c$ $\therefore a^{1729}+b^{1729}+c^{1729} \\ =a+b+c=0$ Since sum of roots of $x^3-1=0$ is $\frac{0}{1}=0$ .

$a^{1729}+b^{1729}+c^{1729} = a^{1728}\cdot a+b^{1728}\cdot b+c^{1728}\cdot c$ $(a^{3})^{576}\cdot a + (b^{3})^{576}\cdot b+(c^{3})^{576}\cdot c=a+b+c$ Using Vieta's Formula we have $a+b+c=0$