If $$a = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + ...\infty$$ $$b = x + \frac{x^4}{4!} + \frac{x^7}{7!} + ...\infty$$ $$c = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + ...\infty$$ show that $$a^3 + b^3 + c^3 -3abc = 1$$
I understand $$a^3 + b^3 + c^3 -3abc = (a + b + c) (a^2 + b^2 + c^2 -ab- bc -ca)$$ and $$(a+b+c) = e^x$$ It means somehow I have to derive other factor to be $e^-x$ and there is where my difficult lies. Any help is appreciated.
$$a+b+c=\exp(x)$$ $$a+\omega b+\omega^2 c=\exp(\omega x)$$ $$a+\omega^2 b+\omega c=\exp(\omega^2 x)$$ where $\omega=\exp(2\pi i/3)$. Thus $$(a+b+c)(a+\omega b+\omega^2 c)(a+\omega^2 b+\omega c)=\exp((1 +\omega+\omega^2)x).$$ This simplifies to $$a^3+b^3+c^3-3abc=1.$$