Factorise completely $x^4 - y^4$. Answer ... Given that $x^2 - 2xy = 1 - y^2$, express $y$ in terms of $x$. Answer ... or ...

1. Recognise $x^4-y^4$ as a difference of squares.

$x^4-y^4=(x^2)^2-(y^2)^2$

$=(x^2-y^2)(x^2+y^2)$

2. Factorise the remaining difference of squares completely.

$x^2-y^2=(x-y)(x+y)$

So,

$x^4-y^4=(x-y)(x+y)(x^2+y^2)$

1. Collect all terms on one side and write the equation as a quadratic in $y$.

$x^2-2xy=1-y^2$

$y^2-2xy+x^2-1=0$

2. Recognise and factorise the perfect square expression.

$y^2-2xy+x^2-1=0$

$(y-x)^2-1=0$

$[(y-x)-1][(y-x)+1]=0$

3. Solve the two linear factors and express $y$ in terms of $x$.

$(y-x)-1=0$ or $(y-x)+1=0$

$y-x=1$ or $y-x=-1$

$y=x+1$ or $y=x-1$