Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

$p(x)=x^{3}-a x^{2}+6 x-a$

$g(x)=x-a$

By remainder theorem, when $p(x)$ is divided by $(x-a)$, then the remainder $=p(a)$.

Putting $x=a$ in $p(x)$, we get

$p(a)=a^{3}-a \times a^{2}+6 \times a-a=a^{3}-a^{3}+6 a-a=5 a$

$\therefore$ Remainder $=5 a$

Thus, the remainder when $p(x)$ is divided by $g(x)$ is $5 a$.