Intermediate Algebra (12th Edition)

$9m^2-6my+y^2-z^2$
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the given expression, $[(3m-y)+z][(3m-y)-z] ,$ is equivalent to \begin{array}{l}\require{cancel} (3m-y)^2-(z)^2 \\\\= (3m-y)^2-z^2 .\end{array} Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the expression above is equivalent to \begin{array}{l}\require{cancel} [(3m)^2+2(3m)(-y)+(-y)^2]-z^2 \\\\= 9m^2-6my+y^2-z^2 .\end{array}