Let $z=x+iy$  be a complex number where $x$  and $y$  integers. Then the area of the rectangle whose vertices are the roots of the equation:

           $z{\overline{z}}^3+\overline{z}{z}^3=350$  is:

   $\because$        $\overline{z}\left({\overline{z}}^2+z^2\right)=350$ 

Put     $z=x+iy$

$\Rightarrow$        $\left(x^2+y^2\right).2\left(x^2-y^2\right)=350$

$\Rightarrow$    $\left(x^2+y^2\right)\left(x^2-y^2\right)=175=25\times 7$

$\Rightarrow$      $x^2+y^2=25,x^2-y^2=7$

or           $x^2=16,y^2=9$

$\therefore$          $x\pm 4,y=\pm 3,x,y\in I$

    $Area=8\times 6=48sq.units$