Let $\,(a,b) = (d).\,$ Then $\,a,b \in (d)\,\Rightarrow\, d\mid a,b,\,$ so $\,d\,$ is a common divisor of $\,a,b.\,$
Conversely $\,d\in (a,b)\,$ so $\,d = r a + sb,\ r,s\in R,\ $ so $\ c\mid a,b\,\Rightarrow\, c\mid d = ra+sb.\,$
Hence $\,d\,$ is a common divisor of $\,a,b\,$ that is divisible by every common divisor. Therefore, by definition $\,d\,$ is a greatest common divisor of $\,a,b.$
Hint for $(b):\ $ $\,1\in (a,b)\iff (1) = (a,b) = (\gcd(a,b))\ $ by above
This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 7 x + 3 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( 2 x^{2} - 7 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } + \left( \frac{ 15 x - 45 }{ 4 } \right) $$
$$ \left( 2 x^{2} - 7 x + 3 \right) = \left( \frac{ 15 x - 45 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 5 }{ 4 } \right) }{ \left( \frac{ 2 x - 9 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 12 x + 20 }{ 15 } \right) }{ \left( \frac{ 4 x^{2} - 20 x + 24 }{ 15 } \right) } $$
$$ \left( x^{2} - 3 x + 5 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{2} - 5 x + 6 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( -1 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{2} - 3 x + 5 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( x^{2} - 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( - x + 3 \right) $$
............
Best Answer
Remember that $d$ is a greatest common divisor of $a$ and $b$ if and only if:
Let $a$ and $b$ be integers, and let $d$ be their greatest common divisor as integers. Then we know that $d$ satisfies the two properties above, where "divides" means "divides in $\mathbb{Z}$; and $c$ in point 2 is an arbitrary integer.
The thing that makes this very easy is that in the integers, we can express a gcd as a linear combination: we know that there exist integers $\alpha$ and $\beta$ such that $d=\alpha a + \beta b$.
Thus, if $\mathbf{x}$ is a Gaussian integer that divides $a$ and divides $b$ (in the Gaussian integers), then it also divides $\alpha a$, it also divides $\beta b$, and hence also divides $\alpha a + \beta b$. Thus, if $\mathbf{x}$ divides $a$ and divides $b$ in the Gaussian integers, then it divides $d$ in the Gaussian integers.
Thus, $d$ is a greatest common divisor for $a$ and $b$ in the Gaussian integers, since it satisfies the two requirements to be a greatest common divisor:
Thus, $d$ is a greatest common divisor for $a$ and $b$ in $\mathbb{Z}[i]$.
No need to muck about with norms or with products.
P.S. Note that in the integers we can talk about "the" greatest common divisor because, even though every pair of numbers (except for $0$ and $0$) has two greatest common divisors ($d$ and $-d$), there is a natural way to "choose" one of them. In the Gaussian integer each pair of numbers (again, except $0$ and $0$) has four greatest common divisors (if $d$ is one, then so are $-d$, $id$, and $=id$); there is no universally accepted way of deciding which one would be "the" greatest common divisor, so generally you should talk about a greatest common divisor rather than the greatest common divisor.