Taylor series equation Let $f(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$ and $g(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n+1}}{(3n+1)!}$ and $h(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n+2}}{(3n+2)!}$ Show that $f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=1.$  Today a calculus student asked me this question.  first thing that came in my mind that it is not true since if you take x=0 you will get 0=1, but someone pointed out to me that $f(0) $ will have $0^0$. Finally, finally I think I managed to solve it but under the assumptoin that $f(0)=1$, can you help to solve this question without any further assumptions.Hint: may be it is usefull to notice that $h'=g$ and $g'=f$ and $f'=h$.
 A: The problem probably relies on the less known identity
$$a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) $$
Thus, we need to show
$$f^2+g^2+h^2-fg-fh-gh=\frac{1}{f+g+h}=e^{-x}$$
Now, let $u(x)=f^2+g^2+h^2-fg-fh-gh$. Then
$$u'=2ff'+2gg'+2hh'-f'g-fg'-fh'-f'h-gh'-g'h$$
$$=2fh+2gf+2hg-hg-ff-fg-hh-gg-fh=-u$$
this proves that 
$$u(x)=Ce^{-x}$$
for some constant $C$. 
Combining this we get: there exists a constant $C$ so that  
$$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=C$$
Now setting $x=0$ we get $C=f(0)^3$ which seems to complete the proof.
I think that one can argue that this solution works over $\mathbb C$, and then setting $x=\omega$ a primitive third root of unity leads directly to $C=1$, which emphasizes that $f(0)=1$ is the only natural choice... 
P.S.
$f(0)=1$ is not an assumption. the definition of the Taylor series is
$$f(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}=1+\frac{x^{3}}{(3)!}+\frac{x^{6}}{(6)!}+...$$
Note that this is the same as
$$e^x= \sum_{n=0}^{\infty}\frac{x^{n}}{(n)!}$$
What is $e^0$?   
A: Easliy you can show that 
$$g'(x)=f(x) \tag 1$$
$$h'(x)=g(x) \tag 2$$
$$f'(x)=h(x) \tag 3$$
$U(x)=f(x)g(x)h(x)$
$U'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$
$U'(x)=h^2(x)g(x)+f^2(x)h(x)+g^2(x)f(x)$

$P(x)=f^3(x)+g^3(x)+h^3(x)$
$P'(x)=3(f^2(x)f'(x)+g^2(x)g'(x)+h^2(x)h'(x))$
$P'(x)=3(f^2(x)h(x)+g^2(x)f(x)+h^2(x)g(x))$
You can see that 
$3U'(x)=P'(x)=3(h^2(x)g(x)+f^2(x)h(x)+g^2(x)f(x))$
$3U(x)=P(x)+c$  
where c is a constant.
$3f(x)g(x)h(x)=f^3(x)+g^3(x)+h^3(x)+c$
We know that for $x=0$ 
$f(0)=1$,$g(0)=0$,$h(0)=0$
$c=-1$
Thus
$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=1$
A: Differentiate the given expression with respect to $x$ and use the cyclic nature of the derivatives noted in the question. We obtain
$$3f^2f'+3g^2g'+3h^2h'-3(ghf'+fg'h+fgh').$$
Since $f'=h$, $g'=f$, and $h'=g$, this is zero. So the expression is a constant, and setting $x=1$ shows this constant is $1$. 
