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

Explanation:

Step1: Recall the difference of squares formula

The difference of squares formula is ( (a + b)(a - b)=a^{2}-b^{2} ), which means we need two binomials where one is a sum and the other is a difference of the same two terms.

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

Rewrite ((y - x)) as (-(x - y)), so ((x - y)(y - x)=(x - y)\times[-(x - y)]=-(x - y)^{2}), which is a perfect square trinomial (after expanding), not a difference of squares.

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

This is ((6 - y)^{2}), a perfect square trinomial, not a difference of squares (since both binomials are the same, it's a square of a binomial, not a product of a sum and a difference).

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

Rewrite it as ((xz + 3)(xz - 3)), which fits the difference of squares formula with (a = xz) and (b = 3). So ((xz + 3)(xz - 3)=(xz)^{2}-3^{2}=x^{2}z^{2}-9), a difference of squares.

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

This fits the difference of squares formula with (a = y^{2}) and (b = xy). So ((y^{2}-xy)(y^{2}+xy)=(y^{2})^{2}-(xy)^{2}=y^{4}-x^{2}y^{2}), a difference of squares.

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

Rewrite it as ((64y^{2}+x^{2})(64y^{2}-x^{2})), which fits the difference of squares formula with (a = 64y^{2}) and (b = x^{2}). So ((64y^{2}+x^{2})(64y^{2}-x^{2})=(64y^{2})^{2}-(x^{2})^{2}=4096y^{4}-x^{4}), a difference of squares.

Answer:

((3 + xz)(- 3+xz)), ((y^{2}-xy)(y^{2}+xy)), ((64y^{2}+x^{2})(-x^{2}+64y^{2})) (corresponding to the third, fourth, and fifth options)