Causal effect by back-door and front-door adjustments If we wanted to calculate the causal effect of $X$ on $Y$ in the causal graph below, we can use both the back-door adjustment and front-Door adjustment theorems, i.e.,
$$P(y | \textit{do}(X = x)) = \sum_u P(y | x, u) P(u)$$
and 
$$P(y | \textit{do}(X = x)) = \sum_z P(z | x) \sum_{x'} P(y|x', z)P(x').$$
Is it an easy homework to show that the two adjustments lead to the same causal effect of $X$ on $Y$? 

 A: The action $do(x)$ corresponds to an intervention on variable $X$ that sets it to $x$.  When we intervene on $X$, this means the parents of $X$ do not affect its value anymore, which corresponds to removing the arrows pointing to $X$.So let's represent this intervention on a new DAG.

Let's call the original observational distribution $P$  and the post-intervention distribution $P^*$. Our goal is to express $P^*$ in terms of $P$. Notice that in $P^*$ we have that $U \perp X$. Also, the pre interventional and post interventional probabilities share these two invariances: $P^*(U) = P(U)$ and $P^*(Y|X, U) = P(Y|X, U)$ since we did not touch any arrow entering those variables in our intervention. So:
$$
\begin{aligned}
P(Y|do(X)) &:= P^*(Y|X)  \\
           &=\sum_{U}P^*(Y|X, U)P^*(U|X)\\
           &=\sum_{U}P^*(Y|X, U)P^*(U)\\
           &=\sum_{U}P(Y|X, U)P(U)
\end{aligned}
$$
The derivation of the front door is a bit more elaborate. First notice that there's no confounding between $X$ and $Z$, hence,
$$P(Z|do(X)) = P(Z|X)$$ 
Also, using the same logic for deriving $P(Y|do(X))$ we see that controlling for $X$ is enough for deriving the effect of $Z$ on $Y$, that is 
$$P(Y|do(Z)) = \sum_{X'}P(Y|X', Z) P(X')$$
Where I'm using the prime for notation convenience for the next expression. So these two expressions are already in terms of the pre-intervention distribution, and we simply used the previous backdoor rationale to derive them. 
The last piece we need is to infer the effect of $X$ on $Y$ combining the effect of $Z$ on $Y$ and $X$ on $Z$. To do that, notice in our graph $P(Y|Z, do(X)) = P(Y|do(Z), do(X)) = P(Y|do(Z))$, since the effect of $X$ on $Y$ is completely mediated by $Z$ and the backdoor path from $Z$ to $Y$ is blocked when intervening on $X$. Hence:
$$
\begin{aligned}
P(Y|do(X))  &= \sum_{Z} P(Y|Z, do(X))P(Z|do(X))\\
            &= \sum_{Z} P(Y|do(Z))P(Z|do(X))\\
            &= \sum_{Z} \sum_{X'}P(Y|X', Z) P(X')P(Z|X)\\
            &= \sum_{Z}P(Z|X) \sum_{X'}P(Y|X', Z) P(X')
\end{aligned}
$$
Where $\sum_{Z} P(Y|do(Z))P(Z|do(X))$ can be understood in the following way: when I intervene on $Z$, then the distribution of $Y$ changes to $P(Y|do(Z))$; but I'm actually intervening on $X$ so I want to know how often would $Z$ take a specific value when I change $X$, which is $P(Z|do(X))$.
Hence, the two adjustments give you the same post-interventional distribution on this graph, as we have showed.

Re-reading your question it occurred to me you might be interested in directly showing that the right hand side of the two equations are equal in the pre-interventional distribution (which they must be, given our previous derivation). That's not hard to show directly too. It suffices to show that in your DAG:
$$
\sum_{X'}P(Y|Z, X') P(X') = \sum_{U}P(Y|Z, U) P(U)
$$
Notice the DAG implies $Y \perp X|U, Z$ and $U \perp Z|X$ then:
$$
\begin{aligned}
\sum_{X'}P(Y|Z, X') P(X') &= \sum_{X'}\left(\sum_{U}P(Y|Z, X', U)P(U|Z, X') \right)P(X') \\
&= \sum_{X'}\left(\sum_{U}P(Y|Z,  U)P(U| X') \right)P(X') \\
&= \sum_{U}P(Y|Z,  U) \sum_{X'}P(U| X')P(X') \\
&= \sum_{U}P(Y|Z,  U) P(U) \\
\end{aligned}
$$
Hence:
$$
\begin{aligned}
\sum_{Z}P(Z|X) \sum_{X'}P(Y|X', Z) P(X') &= \sum_{Z}P(Z|X)\sum_{U}P(Y|Z,  U) P(U)\\
&= \sum_{U}P(U)\sum_{Z}P(Y|Z, U)P(Z|X) \\
&= \sum_{U}P(U)\sum_{Z}P(Y|Z, X, U)P(Z|X, U) \\
&= \sum_{U}P(Y| X, U) P(U)\\
\end{aligned}
$$
