Answer
$y^{5}+4y^{4} +4y^{3}-y$
Work Step by Step
If we factor out y from the first parentheses,
$y(y^{2}+2y+1)(y^{2}+2y-1)=...,$
we may recognize a special product, a difference of squares:
... $I^{2}-II^{2}=(I+II)(I-II)$
$=y[(y^{2}+2y)+1]\cdot[(y^{2}+2y)-1]=y[(y^{2}+2y)^{2}-1^{2}]$
... first term in the brackets is a square of a sum ...
... $(I+II)^{2}= I^{2}+2\cdot I\cdot II+II^{2}$,
$=y[(y^{2})^{2}+2(y^{2})(2y)+(2y)^{2}-1]$
$= y(y^{4}+4y^{3}+4y^{2}-1)$
$=y^{5}+4y^{4} +4y^{3}-y$
