JAMB 1989 · UME · Q19

Factorize completely $y^3 - 4xy + xy^3 - 4y$ .

Worked solution (try it first)

Group the terms with $4y$ and $4xy$ : $(y^3 - 4y) + (xy^3 - 4xy)$ .
Take out $y$ from the first group and $xy$ from the second: $y(y^2 - 4) + xy(y^2 - 4)$ .
Take out the common bracket and then $y$ : $y(1 + x)(y^2 - 4)$ .
Difference of two squares: $y^2 - 4 = (y + 2)(y - 2)$ , so you get $y(1 + x)(y + 2)(y - 2)$ , option C.