Prove that
$$ \gcd (a, b, c) = \gcd (a, \gcd (b, c)) $$
where $a, b, c$ are non-zero.
As shorthand, let $f=\gcd(a,b,c)$, $g=\gcd(a,\gcd(b,c))$ and $h=\gcd(b,c)$. Our aim is to show $f=g$.
By definition of gcd, we have
$$ g\mid a \quad \quad g \mid h $$
$$ h \mid b \quad \quad h \mid c $$
Since $g \mid h$ and $h \mid b$, we have $g \mid b$, by Theorem 1.2. Similarly, since $g \mid h$ and $h \mid c$, we have $g \mid c$.
In summary
$$ g\mid a \quad \quad g \mid b \quad \quad g \mid c $$
This tells us $g$ is a common divisor of $a$, $b$ and $c$, but does not tell us it is the greatest common divisor like $f$ is, so
$$g \leq f$$
We also have
$$ f\mid a \quad \quad f \mid b \quad \quad f \mid c $$
The result from exercise (1.3).18 tells us
$$h = \gcd (b, c) \quad \land \quad f\mid b \quad \land \quad f \mid c \quad \implies \quad f \mid h $$
So we have
$$ f \mid a \quad f \mid \gcd (b, c)$$
This tells is $f$ is a common divisor of $a$ and $gcd(b,c)$, but does tell us it is the greatest common divisor like $g$ is, so
$$ f \leq g $$
Both $g \leq f$ and $f \leq g$ means $f=g$, that is $ \gcd (a, b, c) = \gcd (a, \gcd (b, c)) $.