If X, Y, Z are independent random variables, then X + Y, Z are independent random variables. I found the same question (X,Y,Z are mutually independent random variables. Is X and Y+Z independent? here), but the answer uses characteristic functions and fourier inversion theorem, but this is exercise in chapter long before characteristic functions.
 A: The answer in the link above using the concept of sigma-algebra is the most general approach and is the best. If you have not studied that yet, here I provide another approach involving the CDF and law of total probability.
Note that $X, Y, Z$ are mutually independent if and only if
$$ \Pr\{X \leq x, Y \leq y, Z \leq z\} = \Pr\{X \leq x\}\Pr\{Y \leq y\}\Pr\{Z \leq z\}$$
for any $x, y, z \in \mathbb{R}$.
Now, for any $w, z \in \mathbb{R}$
$$ \begin{align}\Pr\{X + Y \leq w, Z \leq z\} 
& = \int_{-\infty}^{+\infty} \Pr\{X + Y \leq w, Z \leq z|X = x\}dF_X(x) \\ 
& = \int_{-\infty}^{+\infty} \Pr\{x + Y \leq w, Z \leq z\}dF_X(x) \\ 
& = \int_{-\infty}^{+\infty} \Pr\{x + Y \leq w\}\Pr\{Z \leq z\}dF_X(x) \\ 
& = \Pr\{Z \leq z\}\int_{-\infty}^{+\infty} \Pr\{x + Y \leq w\}dF_X(x) \\ 
& = \Pr\{Z \leq z\}\Pr\{X + Y \leq w\} \\ 
\end{align}$$
where the first and the fifth equalities are using the law of total probability, the second and the third equalities are using the given mutual independence, and the fourth equality is pulling out the term that is independent of the integrating variable $x$. Here $F_X(x) = \Pr\{X \leq x\}$ is the CDF of $X$.
