<s>[INST] Exponential series problem 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. 
 [/INST] $$a+b+c=\exp(x)$$
$$a+\omega b+\omega^2 c=\exp(\omega x)$$
$$a+\omega^2 b+\omega c=\exp(\omega^2 x)$$
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
$$a^3+b^3+c^3-3abc=1.$$
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