A quicker route is to recognize this formula as Simpson's 1/3 rule applied to the interval $[-1,1]$ with nodes $-1, 0, 1$.
Simpson's 1/3 rule over $[x_0,x_2]$ with step $h=(x_2-x_0)/2$ states\[\int_{x_0}^{x_2}f(x)\,dx\approx \frac{h}{3}\left[f(x_0)+4f(x_1)+f(x_2)\right]\]
Here $x_0=-1$, $x_1=0$, $x_2=1$, so $h=\frac{1-(-1)}{2}=1$. Substituting,\[\int_{-1}^1 f(x)\,dx\approx \frac{1}{3}\left[f(-1)+4f(0)+f(1)\right]=\frac{1}{3}f(-1)+\frac{4}{3}f(0)+\frac{1}{3}f(1)\]
Matching coefficients with $af(-1)+bf(0)+cf(1)$ gives $a=\frac{1}{3}$, $b=\frac{4}{3}$, $c=\frac{1}{3}$. Since Simpson's 1/3 rule is a classical result known to integrate cubic polynomials exactly, its error term involves the fourth derivative, this immediately confirms exactness up to degree $3$ without solving the linear system from scratch.
\[\boxed{a=\frac{1}{3},\ b=\frac{4}{3},\ c=\frac{1}{3}}\]