How do I prove that $$ a^3+b^3+c^3−3abc = 1 $$ if $ a = 1+\frac{x^3}{3!}+\frac{x^6}{6!}......$ , $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}....$ , $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}......$ ?
I have tried the following :-
I discovered that $a+b+c=e^x$ and that $a=\frac{d(b)}{dx}$, $b=\frac{d(c)}{dx}$, $c=\frac{d(a)}{dx}$ . But I am unable to proceed further with these information.
If all three expressions are considered as functions of $x \in \Bbb R$ (or $\Bbb C$, if you prefer): $$ a(x) = 1+\frac{x^3}{3!}+\frac{x^6}{6!} + \ldots ,\\ b(x)=x+\frac{x^4}{4!}+\frac{x^7}{7!}+ \ldots ,\\ c(x)=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+ \ldots $$ then (as you noticed) $$ b'(x) = a(x) \,, \\ c'(x) = b(x) \,, \\ a'(x) = c(x) \, . $$ It follows that $$ f(x) = a(x)^3+b(x)^3+c(x)^3−3a(x)b(x)c(x) $$ satisfies $$ f'(x) = 3a(x)^2c(x) + 3b(x)^2a(x)+ 3c(x)^2a(x) - 3 \bigl( b(x)c(x)^2 + c(x)a^2(x) + a(x)b^2(x)\bigr) = 0 \, , $$ i.e. $f$ is constant. Setting $x=0$ shows that $f(x) = 1$ for all $x$.
Remark: As Jack said, the expression can be written as a determinant: $$ f(x) = \begin{vmatrix}a(x)&b(x)&c(x) \\ c(x)&a(x)&b(x) \\ b(x)&c(x)&a(x)\end{vmatrix} $$ and this representation can also be used to compute the derivative (omitting the argument for brevity): $$f' = \begin{vmatrix}a'&b'&c' \\ c&a&b \\ b&c&a\end{vmatrix} + \begin{vmatrix}a&b&c \\ c'&a'&b' \\ b&c&a\end{vmatrix} + \begin{vmatrix}a&b&c \\ c&a&b \\ b'&c'&a'\end{vmatrix} \\ = \begin{vmatrix}c&a&b \\ c&a&b \\ b&c&a\end{vmatrix} + \begin{vmatrix}a&b&c \\ b&c&a \\ b&c&a\end{vmatrix} + \begin{vmatrix}a&b&c \\ c&a&b \\ a&b&c\end{vmatrix} = 0 $$
We have $$a^3+b^3+c^3-3abc = \det\begin{pmatrix}a&b&c \\ c&a&b \\ b&c&a\end{pmatrix}=(a+b+c)(a+\omega b+\omega^2 c)(a+\omega^2 b +\omega c)$$
where $\omega=\exp\frac{2\pi i}{3}$ is a primitive third root of unity. On the other hand
$$ a+b+c = e^x,\quad a+\omega b+\omega^2 c = e^{\omega x},\quad a+\omega^2 b +\omega c = e^{\omega^2 x} $$
and $1+\omega+\omega^2 = 0$, hence $a^3+b^3+c^3-3abc = e^{0x} = 1$.
This is just the Wronskian associated to the differential equation $f'''(x)=f(x)$.
Similarly you may prove that there is a homogeneous polynomial $\sum_{cyc}p(a,b,c,d,e)$ with degree $5$ which vanishes over the fundamental solutions of $f^{(5)}(x)=f(x)$.
It is pretty easy if you use $$ a'=c,b'=a, c'=b.$$ Since \begin{eqnarray} &&(a^3+b^3+c^3-3abc)'\\ &=&3a^2a'+3b^2b'+3c^2c'-3(a'bc+ab'c+abc')\\ &=&3a^2c+3b^2a+3c^2b-3(bc^2+a^2c+ab^2)\\ &=&0, \end{eqnarray} one has $$ a^3+b^3+c^3-3abc\equiv C.$$ Using $$ a(0)=1,b(0)=c(0)=0$$ it is easy to see $C=1$ and hence $$ a^3+b^3+c^3-3abc=1. $$
