При каком значении k сумма кубов корней уравнения kx^2-6kx+2k+3=0 равна 72

$x_{1};x_{2}$ -корни уравнения.
$x_{1}^3+x_{2}^3=(x_{1}+x_{2})(x_{1}^2-x_{1}x_{2}+x_{2}^2)\\ x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}\\\\ x_{1}^3+x_{2}^3=(x_{1}+x_{2})((x_{1}+x_{2})^2-3x_{1}x_{2})\\\\ x_{1}*x_{2}=\frac{2k+3}{k}\\ x_{1}+x_{2}=6\\\\ 6(36-\frac{6k+9}{k})=72\\ k=\frac{1}{2}$