Answer
$(y+2)(y^{2}+y+1)$
Work Step by Step
$(y+1)^{3} + 1$
To factor $(y+1)^{3} + 1$, Express each term as the cube of some monomial. $(y+1)^{3} + 1^{3}$
Then use the formula
$A^{3} + B^{3} = ( A + B )(A^{2} -AB+B^{2})$ for factoring.
$(y+1)^{3} + 1$ $= (y+1+1)[(y+1)^{2}-(y+1)(1)+1^{2}]$
$= (y+2) [(y+1)^{2}-(y+1)(1)+1^{2}]$
$= (y+2) [(y+1)^{2}-(y+1)+1]$
$= (y+2) [(y+1)^{2}-y-1+1]$
$= (y+2) [(y+1)^{2}-y]$
The square of the binomial can be found using the formula,
$(a+b)^{2}=a^{2}+2ab+b^{2}$
$= (y+2) [(y)^{2}+2(y)(1)+1^{2}-y]$
$= (y+2) [y^{2}+2y+1-y]$
Combine like terms.
$= (y+2) (y^{2}+y+1)$
