Answer
$r' = \frac{3}{x^{4}}$
$r'' = \frac{-12}{x^{5}}$
Work Step by Step
$r = \frac{(x-1)(x^{2}+x+1)}{x^{3}}$
(For convenience, $\theta$ is replaced with x.)
$r = \frac{x^{3}- 1}{x^{3}}$
$r' = \frac{x^{3}(3x^{2}) - 3x^{2}(x^{3}-1)}{x^{6}}$
$r' = \frac{3x^{2}}{x^{6}}$
$r' = \frac{3}{x^{4}}$
$r'' = \frac{-12}{x^{5}}$
