$( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$, then the values of $a$ and $b$ are respectively.

Given: $( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$.

To do: To find the value of $a$ and $b$.

Solution:

As given, $( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$

Let $x-2=0$

$\Rightarrow x=2$, Put this value in $x^{3}-4 x^{2}+a x+b$.

$x^{3}-4 x^{2}+a x+b$

$\Rightarrow 2^3-4( 2)^2+a\times 2+b=0$

$\Rightarrow 8-16+2a+b=0$

$\Rightarrow 2a+b-8=0$

$\Rightarrow 2a+b=8$  ....... $( i)$

Again, put $x=2$ value in $x^{3}-a x^{2}+b x+8$

$\Rightarrow 2^3-a( 2)^2+b( 2)+8=0$

$\Rightarrow 8-2a+2b+8=0$

$\Rightarrow -2a+2b=-16$   .......$( ii)$

On adding $( i)$ and $( ii)$,

$2a+b-2a+2b=8-16$

$\Rightarrow 3b=-8$

$\Rightarrow b=-\frac{8}{3}$, put this value in $( i)$

$2a-\frac{8}{3}=8$

$\Rightarrow 2a=8+\frac{8}{3}$

$\Rightarrow 2a=\frac{32}{3}$

$\Rightarrow a=\frac{32}{3}\times\frac{1}{2}$

$\Rightarrow a=\frac{16}{3}$

Thus, $a=\frac{16}{3}$ and $b=-\frac{8}{3}$