To solve this complex integral, we first identify that the integral involves a curve \(C\) which is the boundary of the region \(|x| + |y| = 1\). This describes a diamond shape, which can be parameterized as a closed path surrounding a region \(\{z \in \mathbb{C} : |x| + |y| \leq 1\}\).
The function \(\frac{\sin z}{1 - \cos(z^3)}\) contains singularities where \(1 - \cos(z^3) = 0\), i.e., at points where \(\cos(z^3) = 1\). This happens when \(z^3 = 2\pi n\), leading to \(z = (2\pi n)^{1/3} \omega\), where \(n\) is an integer and \(\omega\) is a cube root of unity.
To find singularities within the diamond \(|x| + |y| \leq 1\), note that the endpoints of the paths for \(2\pi n; n=0, 1, -1\) at the cube roots \((2\pi)^{1/3}\), but we will consider only singularities for \(n=0\) as these are the only ones within the contour \(C\). Hence, singularity is at \(z=0\).
Recognizing the integral's symmetry and the nature of analytic continuation, we evaluate the residue of the integrand at \(z=0\), using:
\[\text{Residue} = \lim_{z \to 0} z \cdot \frac{\sin z}{(z^3 + ...)^1} = \lim_{z \to 0} \frac{\sin z}{z^2(1 - ...)}\]
Using Taylor expansion for \(\sin z \approx z\), and \(1 - \cos(z^3) \approx (z^3)^2/2\), we find:
\[\text{Residue at } z = 0 = \frac{1/6}{1/3!} = \frac{1}{2}\]
Utilizing the residue theorem, \(I = \frac{1}{2\pi i} \cdot 2\pi i \cdot \frac{1}{2} = \frac{1}{2}\).
Thus, the value of \(120I\) is:
\[120I = 120 \cdot \frac{1}{2} = 60\]
The solution of \(120I = 60\) lies within the range (2, 2), confirming the value.