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.
 A: Note that $a = \sinh x - x + 1$ and $$b' = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + \cdots$$ Now we can see that $$e^{w x} = 1 + w x + \frac{w^{2}x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots = b' + wb + w^{2}c$$ and $$e^{w^{2}x} = b' + w^{2}b + wc$$ and $$e^{x} = b' + b + c$$ so that $$b = \frac{w^{2}e^{wx} + we^{w^{2}x} + e^{x}}{3}$$ and $$c = \frac{we^{wx} + w^{2}e^{w^{2}x} + e^{x}}{3}$$
Note that $w$ above denotes a cube root of unity with $w \neq 1$. It is easily seen that if $d = b'$ then we have $$b^{3} + c^{3} + d^{3} - 3bcd= (d + c + b)(d + cw + bw^{2})(d + cw^{2} + bw) = e^{x + wx + w^{2}x} = 1$$ Thus $b^{3} + c^{3} + d^{3} = 1 + 3bcd$. I think the expression for $a$ should be equal to $d = 1 + x^{3}/3! + x^{6}/6! + \cdots$ instead of $1 + x^{3}/3! + x^{5}/5! + \cdots$.
A: 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 $a'=b$ , $a''=c$ ,$a'''=a$
