Example 2 Suppose 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.
Solution: By the geometric meaning of the square root of a complex number, the three points corresponding to the complex numbers must satisfy the equation $(x+a)^{3}=b(a, b$ are complex numbers $)$, then we have $x^{3}+p x+1=(x+a)^{3}-b=x^{3}+$ $3 x^{2} a+3 x a^{2}+a^{3}-b$. From this, we can get $3 a=0,3 a^{2}=p, a^{3}-b=1$, that is, $a=0$, $p=0, b=-1$. Therefore, the original equation is $x^{3}+1=0$. Thus, it is known 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, $S=\frac{3 \sqrt{3}}{4}$.