Solution: According to the mean value theorem, if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $\xi \in (a, b)$ such that \[ f'(\xi) = \frac{f(b) - f(a)}{b - a}. \] Here, $f(x) = 1 - x^2 + x^3$, which is a polynomial and hence is continuous and differentiable on $[-1, 1]$. We need to find $\xi \in (-1, 1)$ such that \[ f'(\xi) = \frac{f(1) - f(-1)}{1 - (-1)}. \] First, we calculate $f(1)$ and $f(-1)$: \begin{align*} & f(1) = 1 - 1^2 + 1^3 = 1 \\ & f(-1) = 1 - (-1)^2 + (-1)^3 = -1. \end{align*} Thus, \[ \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - (-1)}{2} = \frac{2}{2} = 1. \]

Next, we find $f'(x)$: \[ f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} (1 - x^2 + x^3) = -2x + 3x^2. \] We need to solve for $\xi$ such that \begin{align*} f'(\xi ) = 1 & \implies -2\xi + 3\xi^2 = 1 \\ & \implies 3\xi^2 - 2\xi - 1 = 0 \\ & \implies (3\xi +1)(\xi - 1) = 0. \end{align*} Thus, \[ \xi = 1 \quad \text{or} \quad \xi = -\frac{1}{3}. \] Since $\xi$ must be in the interval $(-1, 1)$, we have \[ \textcolor{blue}{\boxed{\xi = -\frac{1}{3}}} \]