Question

which products result in a difference of squares? select three options.
□ (x - y)(y - x)
□ (6 - y)(6 - y)
□ (3 + xz)(-3 + xz)
□ (y² - xy)(y² + xy)
□ (64y² + x²)(-x² + 64y²)

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)