(a) Prove that $\operatorname{gcd}(a, b) | \operatorname{gcd}\left(3 a+b, a^{3}\right)$
Since say $d = gcd(a,b)$ then $d|a$ and $d|b$ this will imply $d|(3a+b)$ and $d|a^3$ and therfore $d$ is divisor of those two.But how to show it is greatest common divisor.
One of the basic properties of $e = \gcd(f,g)$ is that if $d \mid f$ and $d \mid g$, then $d$ is a common divisor, so it must also divide the greatest common divisor, i.e., $d \mid e$. As shown in the answer of if $p\mid a$ and $p\mid b$ then $p\mid \gcd(a,b)$, its definition #$1$ uses this explicitly, with a second definition being provable using Bezout's lemma (basically, if $d = \gcd(a,b)$, then $ax + by = d$ for some integers $x,y$, so if $e \mid a$ and $e \mid b$, then $e \mid ax + by = d$). You can also use the Fundamental theorem of arithmetic to show this using prime factorizations, but I won't go into details about that here.
With this concept, you've basically proven what you're asking to show, i.e., with $d = \gcd(a,b)$, then $d \mid 3a + b$ and $d \mid a^3$, so $\gcd(a,b) \mid \gcd(3a + b, a^3)$.
