Step 1: Write down the sum and product of the zeroes.
For $px^2 + qx + r$ with zeroes $\alpha$ and $\beta$:
\[ \alpha + \beta = -\frac{q}{p}, \qquad \alpha\beta = \frac{r}{p} \]
Step 2: Expand the target expression by distributing, instead of factoring the bracket first.
Start from $\alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2+\beta^2)$, and instead write $\alpha^2 + \beta^2$ as $(\alpha+\beta)^2 - 2\alpha\beta$, so:
\[ \alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha+\beta)^2 - 2(\alpha\beta)^2 \]
This spreads the multiplication out into two separate terms rather than keeping everything inside one bracket.
Step 3: Compute the first term, $\alpha\beta(\alpha+\beta)^2$.
\[ \alpha\beta(\alpha+\beta)^2 = \frac{r}{p}\left(-\frac{q}{p}\right)^2 = \frac{r}{p} \times \frac{q^2}{p^2} = \frac{rq^2}{p^3} \]
Step 4: Compute the second term, $2(\alpha\beta)^2$.
\[ 2(\alpha\beta)^2 = 2\left(\frac{r}{p}\right)^2 = \frac{2r^2}{p^2} \]
Step 5: Subtract the two terms, using a common denominator of $p^3$.
\[ \alpha^3\beta + \beta^3\alpha = \frac{rq^2}{p^3} - \frac{2r^2}{p^2} = \frac{rq^2}{p^3} - \frac{2r^2 p}{p^3} = \frac{rq^2 - 2r^2p}{p^3} \]
Factor out $r$ from the numerator:
\[ \alpha^3\beta + \beta^3\alpha = \frac{r(q^2 - 2pr)}{p^3} \]
Step 6: Final Answer.
The value of $\alpha^3\beta + \beta^3\alpha$ is $\frac{r(q^2-2pr)}{p^3}$, so option (A) is correct.
\[ \boxed{\dfrac{r(q^2 - 2pr)}{p^3}} \]