Given that $x - y$ = 1, what is the value of $ x^4- xy^3 -x^3y - 3x^2y + 3xy^2 + y ^4 $?
My attempt: I noticed that $ -3x^2y + 3xy^2 $ can be factorized and become $-3xy(x - y)$. So that the equation become $ x^4- xy^3 -x^3y - 3xy + y ^4 $. However It is not useful in solving the equation.
Note that\begin{align}x^4- xy^3 -x^3y - 3x^2y + 3xy^2 + y ^4&=(x-y)(x^3-3xy-y^3)\\&=(x-y)\bigl((x-y)(x^2+xy+y^2)-3xy\bigr)\end{align}Therefore, if $x-y=1$,\begin{align}x^4- xy^3 -x^3y - 3x^2y + 3xy^2 + y ^4&=x^2+xy+y^2-3xy\\&=x^2-2xy+y^2\\&=(x-y)^2\\&=1.\end{align}