\(\left(x+y+t\right)^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2+t^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2-\left(x^3+y^3\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right)t+3t^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left[x^2+2xy+y^2+3\left(x+y\right)t+3t^2-x^2+xy-y^2\right]\\ =\left(x+y\right)\left[3\left(x+y\right)t+3xy+3t^2\right]\\ =3\left(x+y\right)\left(xt+yt+xy+t^2\right)\\ =3\left(x+y\right)\left[t\left(x+t\right)+y\left(x+t\right)\right]\\ =3\left(x+y\right)\left[\left(x+y\right)t+xy+t^2\right]\\ =3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow2011=3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow D=\dfrac{3\left(x+y\right)\left(y+t\right)\left(t+x\right)}{\left(x+y\right)\left(y+t\right)\left(t+x\right)}=3\)
