How do you factor the expression $x^3 - x^2y - y^3 + xy^2$ ?

Question

8815 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-01T20:57:02

$x^3 - x^2y - y^3 + xy^2=(x-y)(x^2+y^2)$

$x^3 - x^2y - y^3 + xy^2 = y^3((x/y)^3-(x/y)^2+(x/y)-1)$

but

$z^3-z^2+z-1 =0$ has a root $z = 1$

making

$z^3-z^2+z-1 = (z-1)(b z^2+c z+ d)$

equating the coefficients we find

${ (d-1 = 0), (c - d + 1= 0),( b - c -1= 0), (1 - b = 0) :}$

solving for $b,c,d$

$(b=1,c=0,d=1)$

so

$z^3-z^2+z-1 = (z-1)(z^2+ 1)$

and finally

$x^3 - x^2y - y^3 + xy^2=(x-y)(x^2+y^2)$

How do you factor the expression x^3 - x^2y - y^3 + xy^2x^3 - x^2y - y^3 + xy^2?