I NEED IMMEDIATE HELP!!!!Which products result in a difference of squares? Check all that apply.[A] (x-y)(y-x) [B] (6-y)(6-y) [C] (3+xz)(-3+xz) [D] (y^2-xy)(y^2+xy) [E] (25x-7y)(-7y+25x)[F] (64y^2+x^2)(-x^2+64y^2)

Hello!The answers are:C) [tex](3+xz)(-3+xz)[/tex]D) [tex](y^2-xy)(y^2+xy)[/tex]F) [tex](64y^2+x^2)(-x^2+64y^2)[/tex]Why?To know which of the products results in a difference of square, we need to remember the difference of squares from:The difference of squares form is:[tex](a+b)(a-b)=a^{2}-b^{2}[/tex]So, discarding each of the given options in order to find which products result in a difference of squares, we have:A)[tex](x-y)(y-x)=xy-x^{2}-y^{2} +yx=-x^{2} -y^{2}[/tex]So, the obtained expression is not a difference of squares.B)[tex](6-y)(6-y)=36-6y-6y+y^{2}=y^{2}-12y+36[/tex]So, the obtained expression is not a difference of squares.C)[tex](3+xz)(-3+xz) =(xz+3)(xz-3)=(xz)^{2}-3xz+3xz-(3)^{2}\\\\(xz)^{2}-3xz+3xz-(3)^{2}=(xz)^{2}-(3)^{2}[/tex]So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.D)[tex](y^2-xy)(y^2+xy)=(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2} \\\\(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2}=(y^{2})^{2}-(xy)^{2}[/tex]So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.E)[tex](25x-7y)(-7y+25x)=-175xy+(25x)^{2}+49y^{2}-175xy\\\\-175xy+(25x)^{2}+49y^{2}-175xy=(25x)^{2}+49y^{2}-350xy[/tex]So, the obtained expression is not a difference of squaresF)[tex](64y^2+x^2)(-x^2+64y^2)=(64y^2+x^2)(64y^2-x^2)\\\\(64y^2+x^2)(64y^2-x^2)=(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}\\ \\(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}=(64y^{2})^{2}-(x^{2})^{2}[/tex]So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.Hence, the products that result in a difference of squares are:C) [tex](3+xz)(-3+xz)[/tex]D) [tex](y^2-xy)(y^2+xy)[/tex]F) [tex](64y^2+x^2)(-x^2+64y^2)[/tex]Have a nice day!