which products result in a difference of squares? select three options.\n(x - y)(y - x)\n(6 - y)(6 - y)\n(3…

Explanation:

Step1: Recall difference - of - squares formula

The difference - of - squares formula is $(a + b)(a - b)=a^{2}-b^{2}$.

Step2: Analyze $(x - y)(y - x)$

$(x - y)(y - x)=-(x - y)(x - y)=-(x - y)^{2}$, not a difference of squares.

Step3: Analyze $(6 - y)(6 - y)$

$(6 - y)(6 - y)=(6 - y)^{2}$, not a difference of squares.

Step4: Analyze $(3+xz)(-3 + xz)$

Rewrite it as $(xz + 3)(xz-3)$. Here $a = xz$ and $b = 3$. Using the difference - of - squares formula, $(xz + 3)(xz - 3)=(xz)^{2}-3^{2}=x^{2}z^{2}-9$.

Step5: Analyze $(y^{2}-xy)(y^{2}+xy)$

Here $a = y^{2}$ and $b = xy$. Using the difference - of - squares formula, $(y^{2}-xy)(y^{2}+xy)=(y^{2})^{2}-(xy)^{2}=y^{4}-x^{2}y^{2}$.

Step6: Analyze $(64y^{2}+x^{2})(-x^{2}+64y^{2})$

Rewrite it as $(64y^{2}+x^{2})(64y^{2}-x^{2})$. Here $a = 64y^{2}$ and $b = x^{2}$. Using the difference - of - squares formula, $(64y^{2}+x^{2})(64y^{2}-x^{2})=(64y^{2})^{2}-(x^{2})^{2}=4096y^{4}-x^{4}$.

Answer:

C. $(3+xz)(-3 + xz)$, D. $(y^{2}-xy)(y^{2}+xy)$, E. $(64y^{2}+x^{2})(-x^{2}+64y^{2})$