Answer
$3b^2$
Work Step by Step
By substituting, we get the intermediate form $\frac{0}{0}$. So by factorizing the numerator, we have
$$
\lim _{x \rightarrow b} \frac{ x^3b^3 }{x-b }=\lim _{x \rightarrow b} \frac{ (x -b)(x^2+bx+b^2)}{x-b}\\
=\lim _{x \rightarrow b} x^2+bx+b^2 = b^2+b^2+b^2=3b^2.
$$
