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 $Y$?
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:
$$ \begin{aligned} \sum_{Z}P(Z|X) \sum_{X'}P(Y|X', Z) P(X') &= \sum_{Z}P(Z|X)\sum_{U}P(Y|Z, U) P(U)\\ &= \sum_{U}P(U)\sum_{Z}P(Y|Z, U)P(Z|X) \\ &= \sum_{U}P(U)\sum_{Z}P(Y|Z, X, U)P(Z|X, U) \\ &= \sum_{U}P(Y| X, U) P(U)\\ \end{aligned} $$
