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Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Q}[X]$. Then $\gcd_S (2,X)=1$ and $\gcd_R(2,X)=2$, where both $R$ and $S$ are GCD-domains.

$Edited$: Take $R=\mathbb{Z}+X\mathbb{Q}[X]$ instead of $\mathbb{Q}[X]$, then $2$ is nonunit in $R$ and $2$ divides $X$ in $R$. Hence $\gcd_R(2,X)=2$.

Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Z}+X\mathbb{Q}[X]$. Then $gcd_S (2,X)=1$ and $gcd_R(2,X)=2$, where both $R$ and $S$ are GCD-domains.

Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Z}+X\mathbb{Q}[X]$. Then $gcd_S (2,X)=1$ and $gcd_R(2,X)=2$,

where both $R$ and $S$ are GCD-domains.

$Edited$: Take $R=\mathbb{Z}+X\mathbb{Q}[X]$ instead of $\mathbb{Q}[X]$  , then $2$ is nonuunit in $R$ and $2$ divides $X$ in $R$. Hence $gcd_R(2,X)=2$

Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Q}[X]$. Then $\gcd_S (2,X)=1$ and $\gcd_R(2,X)=2$, where both $R$ and $S$ are GCD-domains.

$Edited$: Take $R=\mathbb{Z}+X\mathbb{Q}[X]$ instead of $\mathbb{Q}[X]$, then $2$ is nonunit in $R$ and $2$ divides $X$ in $R$. Hence $\gcd_R(2,X)=2$.

Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Q}[X]$. Then $gcd_S (2,X)=1$ and $gcd_R(2,X)=2$,

where both $R$ and $S$ are GCD-domains.

$Edit$: Take $R=\mathbb{Z}+X\mathbb{Q}[X]$ instead of $\mathbb{Q}[X]$ , then $2$ is nonuunit in $R$ and $2$ divides $X$ in $R$. Hence $gcd_R(2,X)=2$

Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Z}+X\mathbb{Q}[X]$. Then $gcd_S (2,X)=1$ and $gcd_R(2,X)=2$,

where both $R$ and $S$ are GCD-domains.

$Edited$: Take $R=\mathbb{Z}+X\mathbb{Q}[X]$ instead of $\mathbb{Q}[X]$ , then $2$ is nonuunit in $R$ and $2$ divides $X$ in $R$. Hence $gcd_R(2,X)=2$