Answer:
[tex](3+xz)(-3+xz)[/tex]
[tex](y^{2}-xy)(y^{2}+xy)[/tex]
[tex](64y^{2}+x^{2})(-x^{2}+64y^{2})[/tex]
Step-by-step explanation:
we know that
The difference of squares is equal to
[tex](a+b)(a-b)=a^{2}-b^{2}[/tex]
so
case A) [tex](x-y)(y-x)[/tex]
[tex](x-y)(y-x)=-(x-y)(x-y)=-(x-y)^{2}[/tex] ------> is not a difference of squares
case B) [tex](6-y)(6-y)[/tex]
[tex](6-y)(6-y)=(6-y)^{2}[/tex] -----> is not a difference of squares
case C) [tex](3+xz)(-3+xz)[/tex]
[tex](3+xz)(-3+xz)=xz^{2}-3^{2}[/tex] -----> is a difference of squares
case D) [tex](y^{2}-xy)(y^{2}+xy)[/tex]
[tex](y^{2}-xy)(y^{2}+xy)=(y^{2})^{2}-xy^{2}[/tex] -----> is a difference of squares
case E) [tex](25x-7y)(-7y+25x)[/tex]
[tex](25x-7y)(-7y+25x)=-(25x-7y)^{2}[/tex] ------> is not a difference of squares
case F) [tex](64y^{2}+x^{2})(-x^{2}+64y^{2})[/tex]
[tex](64y^{2}+x^{2})(-x^{2}+64y^{2})=(64y^{2})^{2}-(x^{2})^{2}[/tex] -----> is a difference of squares
