How many stereoisomers are possible for the compound in picture? How many stereoisomers are possible for the compound in picture? I'm getting 8. Is that correct?

My approach was as follows: The left hand methyl groups may be both up or one up and one down. For each of these cases the groups on the right may be in $(E{,}Z'),(E'{,}Z),(E{,}E'),(Z{,}Z')$ configurations. So $4\cdot2=8$.

I think it is quite evident that carbon having 2 &#x3c0; bonds must be linear. I apologize for the non-linearity in the picture. I hope you will be able to understand the question otherwise.
Follow these steps:


Identify all possible sources of stereochemistry. You got pretty far.
Determine how many isomers can derive from each. I think you went wrong here.
Do you have symmetry? Reduce accordingly.
Multiply all the numbers together.


In your molecule:


Double bonds:


the internal double bond in the residue on the top right
the cumulene, which also shows isomerism (E/Z for uneven double bond counts, aR/aS for even ones).


Asymmetric carbons:


the top methyl group
the bottom methyl group
the cumulene-containing side-chain
the other unsaturated side-chain.

Every one of these can be either R/S or E/Z.
The molecule does not have any elements of symmetry, so we cannot reduce.
All we have are factors of 2; two from the double bonds, another four from the asymmetric carbons. This leads us to:
$$n(\text{isomers}) = 2^{(2 + 4)} = 2^6 = 64$$