\[

\begin{aligned}

f(1) &=f(0)+f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)+\frac{1}{6} f^{(3)}(s) \\

f(-1) &=f(0)-f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)-\frac{1}{6} f^{(3)}(t)

\end{aligned}

\]

for some \(s \in(0,1), t \in(-1,0)\). By subtracting the second equation from the first and using the given values of \(f(1), f(-1)\), and \(f^{\prime}(0)\), we obtain

\[

1=\frac{1}{6}\left(f^{(3)}(s)+f^{(3)}(t)\right),

\]

which is the desired result. Note that we made no use of the hypothesis \(f(0)=0\).