If the equation $2{\mathrm{x}}^2+4\mathrm{xy}+7{\mathrm{y}}^2-12\mathrm{x}-2\mathrm{y}+\mathrm{t}=0$, where t is a parameter has exactly one real solution of the form (x, y), then the sum of ${\mathrm{x}}^3+{\mathrm{y}}^3$is equal to____.

$\begin{array}{l}2{\mathrm{x}}^2+4\mathrm{x}(\mathrm{y}-3)+7{\mathrm{y}}^2-2\mathrm{y}+\mathrm{t}=0\\ \mathrm{D}=0\text{ (for one solution) }\end{array}$
$\Rightarrow 16(\mathrm{y}-3{)}^2-8\left(7{\mathrm{y}}^2-2\mathrm{y}+\mathrm{t}\right)=0$
or  $2(\mathrm{y}-3{)}^2-\left(7{\mathrm{y}}^2-2\mathrm{y}+\mathrm{t}\right)=0$
or  $2\left({\mathrm{y}}^2-6\mathrm{y}+9\right)-\left(7{\mathrm{y}}^2-2\mathrm{y}+\mathrm{t}\right)=0$
or  $-5{\mathrm{y}}^2-10\mathrm{y}+18-\mathrm{t}=0$
or  $5{\mathrm{y}}^2+10\mathrm{y}+\mathrm{t}-18=0$
Again D = 0 (for one solution)
$\Rightarrow 100-20(\mathrm{t}-18)=0$
or   $5-\mathrm{t}+18=0$
or   $\mathrm{t}=23$
for   $\mathrm{t}=23; 5{\mathrm{y}}^2+10\mathrm{y}+5=0$
                    $(\mathrm{y}+1{)}^2=0\text{ or }\mathrm{y}=-1$
for 
$\begin{array}{l}\mathrm{y}=-1;2{\mathrm{x}}^2-16\mathrm{x}+32=0\\ {\mathrm{x}}^2-8\mathrm{x}+16=0\\ \mathrm{x}=4\end{array}$
$\therefore {\mathrm{x}}^3+{\mathrm{y}}^3=-1+64=63$