How to show that $E \left[ ( V-E[V|V+U])^2VU \right] =-2E \left[ \left( \text{Var}(V|V+U) \right)^2 \right]$

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Let $V$ and $U$ be independent random variables (but not identical) and let \begin{align*} W=V+U, \end{align*} Is it true that \begin{align} E \left[ ( V-E[V|W])^2 \frac{VU}{2} \right] &= - E \left[ \left(E\left[ ( V-E[V|W])^2 |W \right] \right)^2 \right]\\ &=-E \left[ \left( Var(V|W) \right)^2 \right], \end{align} where $E\left[ ( V-E[V|W])^2 |W \right]= Var(V|W)$?

We can verify that this is true if $V$ and $U$ are standard normal. Since $E[V|W]= \frac{W}{2}$, on the one hand, we have \begin{align} E \left[ ( V-E[V|W])^2 \frac{VU}{2} \right] = E \left[ ( V-\frac{W}{2})^2 \frac{VU}{2} \right] \\ = E \left[ ( \frac{V}{2}-\frac{U}{2})^2 \frac{VU}{2} \right] =-\frac{1}{4}. \end{align}

On the other hand, using that $Var(V|W)= E[V^2|W]-(E[V|W])^2$ it is not difficult to show that $Var(V|W)=1/2$ and we have \begin{align} -E \left[ \left( Var(V|W) \right)^2 \right]=-\frac{1}{4}. \end{align}

There is a possibility that I am missing a constant factor. So, lets assume that both $U$ and $V$ are zero mean and unit variance (but not identically distributed).

Also, what other good examples can we use to see if it is a valid identity?

Thank you.