What's the background of this exercise? I found this interesting exercise on a calculus book (Stewart)
Let
$$
u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdots
$$
$$
v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\cdots
$$
$$
w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots
$$
Show that $$u^3+v^3+w^3-3uvw=1$$ It turns out to be an interesting application of the 3rd root of unity, which greatly simplified the (could be) tedious calculation. I wonder if it has any deeper interpretation. (At least I don't see how to easily generalize it.) Can anybody explain this? (I don't need help on solving this problem.)
 A: The background is the discrete Fourier transform. The LHS of the identity you wrote down is the determinant of a circulant matrix
$$\left[ \begin{array}{ccc} u & v & w \\\ w & u & v \\\ v & w & u \end{array} \right]$$
and the discrete Fourier transform tells you that its eigenvalues are
$$u + v + w, u + \omega v + \omega^2 w, u + \omega^2 v + \omega w$$
where $\omega$ is a primitive third root of unity, and hence that the product of these is the determinant. But 
$$u + \omega^k v + \omega^{2k} w = \exp (\omega^k x)$$
so their product is
$$\exp (x + \omega x + \omega^2 x) = \exp(0) = 1.$$
The natural generalization involves larger circulant matrices and $\exp (\zeta^k x)$ for $\zeta$ a primitive $n^{th}$ root of unity. Note that when $n = 2$ you get the hyperbolic Pythagorean theorem. 
A: (This isn’t a generalisation, but this is perhaps a more elementary way of looking at the original problem.)
Looking at the three power series, each is the derivative of the next one, cycling around:
$$ u = \frac{dv}{dx} \qquad v = \frac{dw}{dx} \qquad w = \frac{du}{dx} \qquad $$
So differentiating the expression given gives:
$\displaystyle \frac{d}{dx} (u^3 + v^3 + w^3 - 3uvw) $
$\displaystyle \qquad = 3u^2\frac{du}{dx} + 3v^2\frac{dv}{dx} + 3w^2\frac{dw}{dx} - 3(\frac{du}{dx}vw + \frac{dv}{dx}uw + \frac{dw}{dx}uv)$
$\displaystyle \qquad = 3u^2v + 3v^2w + 3uw^2 - 3(v^2w + uw^2 + u^2v)$
$\displaystyle \qquad = 0$
So $u^3 + v^3 + w^3 - 3uvw$ is constant with respect to $x$ (since its derivative is always $0$), and evaluating it at any particular value of $x$ gives its value for all $x$.  Now choosing $x=0$, we get $u = 1$, $v=w=0$, and so $u^3 + v^3 + w^3 - 3uvw = 1$.
