So a Newton polynomial is defined by
$$N(x) = a + b(x-x_0) + c(x-x_0)(x-x_1)$$
Solving this for $x \in \{x_0, x_1, x_2\}$ yields three equations, giving expressions for $a, b, c$:
1: \begin{align*} N(x_0) &= a + b(x_0-x_0) + c(x_0-x_0)(x_0-x_1) \\&= a \end{align*}
2: \begin{align*} N(x_1) &= a + b(x_1-x_0) + c(x_1-x_0)(x_1-x_1) \\&= N(x_0) + b(x_1-x_0) \\\implies b &= \frac{N(x_1)-N(x_0)}{x_1-x_0} \end{align*}
3: \begin{align*} N(x_2) &= a + b(x_2-x_0) + c(x_2-x_0)(x_2-x_1) \\ &= N(x_0) + \frac{N(x_1)-N(x_0)}{x_1-x_0}(x_2-x_0) + c(x_2-x_0)(x_2-x_1) \\\implies c &= \frac{\frac{N(x_2)-N(x_0)}{x_2-x_0}-\frac{N(x_1)-N(x_0)}{x_1-x_0}}{x_2-x_1} \end{align*}
So another expression for $N(x)$ would be:
$$ N(x) = N(x_0) + \frac{N(x_1)-N(x_0)}{x_1-x_0}(x-x_0) + \frac{\frac{N(x_2)-N(x_0)}{x_2-x_0}-\frac{N(x_1)-N(x_0)}{x_1-x_0}}{x_2-x_1}(x-x_0)(x-x_1) $$
However, according to the divided difference method, $c$ should instead be: $$ \frac{\frac{N(x_2)-N(x_\color{red}1)}{x_2-x_\color{red}1}-\frac{N(x_1)-N(x_0)}{x_1-x_0}}{x_2-x_\color{red}0} $$
Is there a mistake in my solution, or does the divided differences yield another solution?