Suppose  $x,y,z$ are real numbers such that $x+y+z=0$ and $xy+yz+zx=-3$, the expression  $\left|x^{3}y+y^{3}z+z^{3}x\right|$  is equal to 

Consider the equation whose roots are $x,y,z$:  $(t-x)(t-y)(t-z)=0.$
This gives $t^3-3t-\lambda =0$, where $\lambda =xyz$. Since  $x,y,z$ are roots of this equation, we have $x^3-3x-\lambda =0,y^3-3y-\lambda =0,z^3-3z-\lambda =0$.
Multiplying the first by y, the second by z and the third by x, we obtain
$\begin{array}{rr} & x^{3}y-3xy-\lambda y=0,\\ & y^{3}z-3yz-\lambda z=0,\\ & z^{3}x-3zx-\lambda x=0.\end{array}$

Adding we obtain  
$x^{3}y+y^{3}z+z^{3}x-3(xy+yz+zx)-\lambda (x+y+z)=0$
This simplifies to
$x^{3}y+y^{3}z+z^{3}x=-9.$
(Here one may also solve for y and z in terms of x and substitute these values in $x^{3}y+y^{3}z+z^{3}x$  to get $-9$)