I'm trying to solve an exercise using Law of iterated expectations, but I'm stuck. I'm not sure if I'm on the right track as expanding the RHS more will not result in answer. I believe I'm looking for an expression that looks like, $$E[Y^2]-E[Y]^2+E[\epsilon^2]-E[\epsilon]^2 = V[Y]+V[\epsilon]$$ but not completly sure.
(e) If $Y = E(Y|X) + \epsilon$ and $E(\epsilon|X)$ is constant then $V(Y)=V(E(Y|X))+V(\epsilon))$. \begin{align*} V[Y] &= E[Y^2]-(E[Y])^2 \\ &= E[E[Y^2|X]+\epsilon]-(E[E[Y|X]+\epsilon])^2 \\ &= E[Y^2]+E[\epsilon]-(E[Y+\epsilon])^2 \\ &= E[Y^2]+E[\epsilon]-(E[Y]^2+E[Y\epsilon]+E[\epsilon]^2) \\ \end{align*}
Any guidance or help would be highly appreciated.
Using the identity $$ \operatorname{Var}(U+V)=\operatorname{Var}(U) + \operatorname{Var}(V) + 2\operatorname{Cov}(U,V)\tag1$$
you just need to show that the covariance between $U:=E(Y\mid X)$ and $V:=\epsilon$ is zero. To do so, apply LIE to write $$ E (UV) = E[ E(UV\mid X) ]= E [ U E(V\mid X) ],\tag2 $$ the last equality following from the fact that $U$ is $X$-measurable. But we are told $E(V\mid X)$ is a constant, say $c$. Upon taking expectations we deduce $c=E(V)$. Continuing (2) we get $$ E (UV) = E[ U E(V) ] = E(U) E(V) $$ and we've proved $U$ and $V$ have zero covariance.