The equality about conditional expectation Let $(U_i,V_i)$ be i.i.d samples from a bivariate distribution. Show that $$\mathbb{E}[(1-U^2_1)\cdot(1-U^2_{2})\cdot|V_1-V_2|]=\mathbb{E}\left\{\mathbb{E}[(1-U^2_1)|V_1]\cdot\mathbb{E}[(1-U^2_2)|V_2]\cdot|V_1-V_2| \right\}$$
I have no ideal how to start.Are there some properties of conditional expectation I should use?
Could you give me some hints?

After a second thought and inspired by @Xi'an's answer, the equality indeed holds.  Below I give a measure-theoretic proof.
By the law of iterative expectation,
\begin{align}
 & E[(1 - U_1^2)(1 - U_2^2)|V_1 - V_2|] \\ 
=& E\{E[(1 - U_1^2)(1 - U_2^2)|V_1 - V_2||V_1, V_2]\} \\
=& E\{|V_1 - V_2|E[(1 - U_1^2)(1 - U_2^2)|V_1, V_2]\}.
\end{align}
So the problem reduces to prove
\begin{align}
E[(1 - U_1^2)(1 - U_2^2)|V_1, V_2] = E[(1 - U_1^2)|V_1, V_2]E[(1 - U_2^2)|V_1, V_2].
\end{align}
To this end, it follows by $\sigma(V_1, V_2) \subset \sigma(V_1, U_2, V_2)$ and the tower property that
\begin{align}
 & E[(1 - U_1^2)(1 - U_2^2)|V_1, V_2] \\
=& E\{E[(1 - U_1^2)(1 - U_2^2)|V_1, U_2, V_2]|V_1, V_2\} \\
=& E\{(1 - U_2^2)E[(1 - U_1^2)|V_1, U_2, V_2]|V_1, V_2\} \\
=& E\{(1 - U_2^2)E[(1 - U_1^2)|V_1]|V_1, V_2\} \\
=& E[(1 - U_1^2)|V_1]E[(1 - U_2^2)|V_1, V_2] \\
=& E[(1 - U_1^2)|V_1]E[(1 - U_2^2)|V_2].
\end{align}
The only step needs further justification is $E[(1 - U_1^2)|V_1, U_2, V_2] = 
E[(1 - U_1^2)|V_1]$ (by the same token, $E[(1 - U_2^2)|V_1, V_2] = E[(1 - U_2^2)|V_2]$), which follows from the following more generalized property (this is Corollary 7.3 in Probability Theory:
Independence, Interchangeability, Martingales by Y. Chow and H. Teicher):

Corollary Let $\mathscr{G}_i$ be a $\sigma$-algebra of events, $i = 1, 2, 3$. If
$\sigma(\mathscr{G}_1 \cup \mathscr{G}_3)$ is independent of $\mathscr{G}_2$, then $\mathscr{G}_1$ and $\mathscr{G}_2$  are conditionally independent given $\mathscr{G}_3$ (as well as unconditionally independent).

Identifying $\mathscr{G}_1, \mathscr{G}_2, \mathscr{G}_3$ in the above result with $\sigma(U_1), \sigma(U_2, V_2), \sigma(V_1)$ respectively finishes the proof.

Proof of Corollary. By definition, it suffices to prove for any $G_i \in \mathscr{G}_i, i = 1, 2, 3$, it holds that
\begin{align}
P(G_1 \cap G_2 | \mathscr{G}_3) = P(G_1|\mathscr{G}_3)P(G_2|\mathscr{G}_3).
\end{align}
Since $\mathscr{G}_2$ is independent of $\mathscr{G}_3$, $P(G_2|\mathscr{G}_3) = P(G_2)$.  Hence it suffices to show that
\begin{align}
P(G_1 \cap G_2 | \mathscr{G}_3) = P(G_1|\mathscr{G}_3)P(G_2). \tag{1}
\end{align}
For any $G_3 \in \mathscr{G}_3$, we have
\begin{align}
 & \int_{G_3}P(G_1|\mathscr{G}_3)P(G_2)dP \\
=& P(G_2)\int_{G_3}P(G_1|\mathscr{G}_3)dP \\ 
=& P(G_2)P(G_1 \cap G_3) \tag{by $G_3 \in \mathscr{G}_3$} \\ 
=& P(G_1 \cap G_3 \cap G_2). \tag{by independence}
\end{align}
This shows $(1)$ holds. Hence the corollary.
