3. If $p$ is a real constant such that the roots of the equation $x^{3}-6 p x^{2}+5 p x+88=0$ form an arithmetic sequence, find $p$.
Solution: Let the roots be $b-d, b$ and $b+d$. From Vieta's formulas,
$$
\begin{aligned}
-88 & =(b-d) b(b+d)=b\left(b^{2}-d^{2}\right) \\
5 p & =(b-d) b+b(b+d)+(b+d)(b-d)=3 b^{2}-d^{2} \\
6 p & =(b-d)+b+(b+d)=3 b
\end{aligned}
$$

From (3), $b=2 p$. Using this on (1) and (2) yields $-44=p\left(4 p^{2}-d^{2}\right)$ and $5 p=$ $12 p^{2}-d^{2}$. By solving each equation for $d^{2}$ and equating the resulting expressions, we get $4 p^{2}+\frac{44}{p}=12 p^{2}-5 p$. This is equivalent to $8 p^{3}-5 p^{2}-44=0$. Since $8 p^{3}-5 p^{2}-44=(p-2)\left(8 p^{2}+11 p+22\right)$, and the second factor has negative discriminant, we only have $p=2$.