Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 18

Take the derivative of $\displaystyle F(x) = \frac{x^3 + 27}{x + 3}$: first, use the Quotient Rule;
then, by dividing the expression before differentiating. Compare your results as a check.
By using Quotient Rule,

$
\begin{equation}
\begin{aligned}
F'(x) &= \frac{(x + 3) \cdot \frac{d}{dx} (x^3 + 27) - (x^3 + 27) \cdot \frac{d}{dx} (x + 3)}{(x + 3)^2}\\
\\
&= \frac{(x + 3)(3x^2) - (x^3 + 27) (1)}{x^2 + 6x + 9}\\
\\
&= \frac{3x^3 + 9x^2 - x^3 - 27}{x^2 + 6x + 9}\\
\\
&= \frac{2x^3 + 9x^2 - 27}{x^2 + 6x + 9}
\end{aligned}
\end{equation}
$


By further simplifying, we get
$F'(x) = 2x - 3$




By dividing the expression first,



Thus,
$\displaystyle F(x) = \frac{x^3 + 27}{x + 3} = x^2 - 3x + 9$

Hence,
$\displaystyle F'(x) = \frac{d}{dx} (x^2 - 3x + 9) = 2x - 3$
Both results agree.