8. It is known that the equation $x^{3}+7 x^{2}+14 x-p=0$ has three distinct roots, and these roots form a geometric progression. Find p and solve this equation.
Answer. $p=-8$; roots of the equation: $x=-1, x=-2, x=-4$.

Solution Let $x, k x, k^{2} x$ be the roots of the given equation, with $x \neq 0, k \neq 0, k \neq \pm 1$ (since the roots are distinct). From Vieta's theorem, it follows that

$$
\left\{\begin{array}{l}
x+k x+k^{2} x=-7 \\
k x^{2}+k^{2} x^{2}+k^{3} x^{2}=14 \\
k^{3} x^{3}=p
\end{array}\right.
$$

Dividing the second equation of this system by the first, we get $k x=-2$. Then the third equation gives $\mathrm{p}=-8$. The original equation becomes $x^{3}+7 x^{2}+14 x+8=0$, from which $(x+2)\left(x^{2}-2 x+4\right)+7 x(x+2)=0,(x+2)\left(x^{2}+5 x+4\right)=0$. Therefore, $x=-1, x=-2$ or $x=-4$.