Let $p(x)=69+11 x-x^{2}+x^{3}, g(x)=x+3$.

$g(x)=x+3=0$ gives $x=-3$

$g(x)$ will be a factor of $p(x)$ if $p(-3)=0 \quad$ (Factor theorem)

Now, $\quad p(-3)=69+11(-3)-(-3)^{2}+(-3)^{3}$

$=69-33-9-27$

$=0$

Since, $p(-3)=0,$ So $g ( x )$ is a factor of $p ( x )$