Answer
$y(y^{2}+1)(y^{4}-y^{2}+1)$
Work Step by Step
$y^{7} + y$
Takeout common factor y
$y^{7} + y = y (y^{6} + 1)$
To factor $(y^{6} + 1)$, express each term as the cube of monomial.
$ = y(y^{6} + 1)$
$ = y((y^{2})^{3} + 1^{3})$
The cube of the binomial can be found using the formula,
$A^{3} + B^{3} = ( A + B )(A^{2} -AB+B^{2})$
$=y [(y^{2}+1)((y^{2})^{2}- (y^{2})(1)+1) ]$
$=y [(y^{2}+1)(y^{4}- y^{2}+1) ]$
$=y (y^{2}+1)(y^{4}- y^{2}+1)$
