## College Algebra (6th Edition)

$(y+2)(y^{2}+y+1)$
$(y+1)^{3} + 1$ 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. $= (y+2) (y^{2}+y+1)$