How do I find the above mentioned moment of inertia?
Steps I've tried:
1.) Triple integrations that proved to be to big.
2.) I noticed that the if we split a $2\times 2\times 2$ into individual $1\times1\times1$ components, the body diagonal of the $2\times 2\times 2$ either passes through or is parallel to body diagonals of the $1\times 1\times 1$ cubes.
If the moment of inertia of the $1\times 1\times 1$ about body diagonal be $I$, then the moment if inertia of the $2\times 2\times 2$ about its body diagonal will be $8I$ because it has 8 times as much mass. I though I could get an equation in terms of $I$ by equating $8I$ and moment of inertia of the individual $1\times 1\times 1$ cubes about the body diagonal of the $2\times 2\times 2$ using parallel axis theorem but it turns out to be greater than $8I$. Could someone resolve my mistake?
P.S: I am not familiar with the concept of tensors.