= Solution
Choose $F(z)=(z+1)^{1/3}z^{1/3}(z-1)^{1/3}$, each factor initially defined by polar argument in $(-\pi,\pi)$. The products of the boundary values agree on $(-\infty,-1)$, so their apparent cuts there cancel; the product extends analytically to $\mathbb C\setminus[-1,1]$ and has $F(z)\sim z$ at infinity. In polar coordinates about the three <branch points> this choice is $\prod r_j^{1/3}\exp[i\sum\theta_j/3]$.
For $0<x<1$, its upper and lower bank values are $e^{\pm i\pi/3}[x(1-x^2)]^{1/3}$. For $-1<x<0$ they are $e^{\pm2i\pi/3}[(-x)(1-x^2)]^{1/3}$. The upper-minus-lower jump on each half is therefore $2i\sin(\pi/3)$ times the positive real integrand. Deforming a positively oriented large circle onto the cut, with the upper bank traversed right to left, gives
$$
\oint F(z)dz=-4i\sin(\pi/3)I.
$$
Small circles about the <branch points> contribute zero in the limit. At infinity,
$$
F(z)=z(1-z^{-2})^{1/3}=z-\frac1{3z}+O(z^{-3}),
$$
so the large-circle integral is $-2\pi i/3$. Hence
$$
\boxed{I=\frac{\pi}{6\sin(\pi/3)}}.
$$
Both the branch choice and the orientation are essential to the sign.
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