13. 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. Find the area of this triangle.
13. By the geometric meaning of the square root of a complex number, the three points corresponding to the complex numbers must satisfy the equation $(x+a)^{3}=b(a, b$ being complex numbers $)$, then we have $x^{3}+$ $p x+1=(x+a)^{3}-b=x^{3}+3 a x^{2}+3 a^{2} x+a^{3}-b$.

From this, we get $3 a=0,3 a^{2}=p, a^{3}-b=1$, which means $a=0, p=0, b=-1$. Therefore, the original equation is $x^{3}+1=0$, from which we can see that the area of the equilateral triangle corresponding to these three points is the area of an equilateral triangle inscribed in a circle with radius 1, which is $\frac{3 \sqrt{3}}{4}$.