$u,v \text{ and } w$

Let $\alpha$ be a primitive cube root of unity. Then $\begin{aligned} u + v + w &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots = \sum_{n = 0}^\infty \frac{x^n}{n!} = e^x, \\ u + \alpha v + \alpha^2 w &= 1 + \alpha x + \alpha^2 \cdot \frac{x^2}{2!} + \frac{x^3}{3!} + \alpha \cdot \frac{x^4}{4!} + \alpha^2 \cdot \frac{x^5}{5!} + \dots = \sum_{n = 0}^\infty \frac{(\alpha x)^n}{n!} = e^{\alpha x}, \\ u + \alpha^2 v + \alpha w &= 1 + \alpha^2 x + \alpha \cdot \frac{x^2}{2!} + \frac{x^3}{3!} + \alpha^2 \cdot \frac{x^4}{4!} + \alpha \cdot \frac{x^5}{5!} + \dots = \sum_{n = 0}^\infty \frac{(\alpha^2 x)^n}{n!} = e^{\alpha^2 x}. \end{aligned}$

Multiplying, we get $(u + v + w)(u + \alpha v + \alpha^2 w)(u + \alpha^2 v + \alpha w) = e^{x + \alpha x + \alpha^2 x}.$ This reduces to $u^3 + v^3 + w^3 - 3uvw = 1.$

I notice that "Nakkache" became "Wehbi". Is there a story behind that?

Hey there, Hana! This particular series problem actually appeared in the Purdue University "Problem Of The Week" publication (GO BOILERS!) I was one of the listed solvers back in Fall 2011.....thanks for the memories!

A detailed solution is worked out here:

Purdue POW - Solution #4 F11