5. Let the three roots of the cubic equation $x^{3}+p x+1=0$ correspond to points in the complex plane that form an equilateral triangle. Then $p=$ , and the area of this equilateral triangle is $\qquad$ .
5.
$p=0, S=3 \sqrt{4}$.
From the definition of coordinate translation and the square root of a complex number, it is known that the complex numbers corresponding to these three points must satisfy the equation $(x+a)^{3}=b$ (where $a, b$ are complex numbers). From $x^{3}+p x+1=(x+a)^{3}-b=x^{3}+3 x^{2} a + 3 x a^{2} + c^{3}-b$, we have $3 a=0, 3 a^{2}=p, a^{3}-b=1$, which means $a=0$, $p=0$, $b=-1$. Therefore, the original equation is $x^{3}+1=0$, with solutions $x_{1}=-1$, $x_{2,3}=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$. From $x_{1}+x_{2}+x_{3}=0$, we know that the center of this equilateral triangle is at the origin, and from $\left|x_{1}\right|=\left|x_{2}\right|=\left|x_{3}\right|=1$ we get
$$
S=\frac{3 \sqrt{3}}{4} .
$$