2. Polynomial
$$
p(x)=x^{3}-224 x^{2}+2016 x-d
$$

has three roots that form a geometric progression. Then the value of $d$ is $\qquad$
2. 729 .

Let the three roots of the polynomial be $a$, $b$, and $c$, and $b^{2}=a c$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b+c=224, \\
a b+b c+c a=2016, \\
a b c=d .
\end{array}\right.
$$

Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$.
Thus, $d=a b c=b^{3}=729$.