由已知条件得:3a=b+d+e,3b=a+c+f,3c=b+d+g,3d=a+c+h;
由于3a+3b+3c+3d=b+d+e+a+c+f+a+c+h=3(a+b+c+d)=2(a+b+c+d)+(e+f+g+h).
所以(a+b+c+d)=e+f+g+h;
则(a+b+c+d)-(e+f+g+h)=0.
由已知条件得:3a=b+d+e,3b=a+c+f,3c=b+d+g,3d=a+c+h;
由于3a+3b+3c+3d=b+d+e+a+c+f+a+c+h=3(a+b+c+d)=2(a+b+c+d)+(e+f+g+h).
所以(a+b+c+d)=e+f+g+h;
则(a+b+c+d)-(e+f+g+h)=0.