<s>[INST] Prove that $a^3+b^3+c^3=1+3abc$, where $a$, $b$ and $c$ are power series. Let $a=1+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\ldots$, $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\ldots$ and $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\ldots$ I want to show that $a^3+b^3+c^3=1+3abc$.
I tried to identify which functions $a$, $b$ and $c$ represent. For instance, $a=\sinh x-x+1$, but I don't know who $b$ and $c$ are. I also thought about using the equality $a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-ac-bc)(a+b+c)$, but obtained nothing.
I would like you to give me a hint to solve this question.
 [/INST] HINT:
By using kth term formula
\begin{align*}a=\frac 1 3 (e^x+e^{\omega \,x}+e^{\omega^2 \,x})\end{align*}
\begin{align*}b=\frac 1 3 (e^x+\omega e^{\omega \,x}+\omega^2e^{\omega^2 \,x})\end{align*}
\begin{align*}c=\frac 1 3 (e^x+\omega^2e^{\omega \,x}+\omega e^{\omega^2 \,x})\end{align*}
Where $\omega^3 -1=0$
You can note that:
$a+b+c=e^x$
and $a'=b$ , $a''=c$ ,$a'''=a$
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