To determine which products result in a difference of squares, let's examine the given expressions one by one.
Recall that a difference of squares is an algebraic expression in the form:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
Let's examine each expression:
1. [tex]\((x-y)(y-x)\)[/tex]
Observe that [tex]\((y-x) = -(x-y)\)[/tex]. Thus:
[tex]\[ (x-y)(y-x) = (x-y)(-(x-y)) = -(x-y)^2 \][/tex]
Since this is a negative square term and not of the form [tex]\(a^2 - b^2\)[/tex], it is not a difference of squares.
2. [tex]\((6-y)(6-y)\)[/tex]
[tex]\[ (6-y)(6-y) = (6-y)^2 \][/tex]
This is a square term and not of the form [tex]\(a^2 - b^2\)[/tex], so it is not a difference of squares.
3. [tex]\((3 + xz)(-3 + xz)\)[/tex]
Recognize that this fits the form [tex]\(a^2 - b^2\)[/tex]:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = xz \][/tex]
[tex]\[ (3 + xz)(-3 + xz) = (xz)^2 - 3^2 = (xz)^2 - 9 \][/tex]
Thus, this is a difference of squares.
4. [tex]\((y^2 - xy)(y^2 + xy)\)[/tex]
This follows the form [tex]\(a^2 - b^2\)[/tex] where:
[tex]\[ a = y^2 \][/tex]
[tex]\[ b = xy \][/tex]
[tex]\[ (y^2 - xy)(y^2 + xy) = (y^2)^2 - (xy)^2 = y^4 - x^2 y^2 \][/tex]
Hence, this is a difference of squares.
5. [tex]\((64y^2 + x^2)(-x^2 + 64y^2)\)[/tex]
Recognize that this follows the form [tex]\(a^2 - b^2\)[/tex] where:
[tex]\[ a = 64y^2 \][/tex]
[tex]\[ b = x^2 \][/tex]
[tex]\[ (64y^2 + x^2)(-x^2 + 64y^2) = (64y^2)^2 - (x^2)^2 = 4096y^4 - x^4 \][/tex]
Therefore, this is also a difference of squares.
Combining all our results, the products that result in a difference of squares are:
1. [tex]\((3 + xz)(-3 + xz)\)[/tex]
2. [tex]\((y^2 - xy)(y^2 + xy)\)[/tex]
3. [tex]\((64y^2 + x^2)(-x^2 + 64y^2)\)[/tex]
So, the three selected options are:
- [tex]\((3 + xz)(-3 + xz)\)[/tex]
- [tex]\((y^2 - xy)(y^2 + xy)\)[/tex]
- [tex]\((64y^2 + x^2)(-x^2 + 64y^2)\)[/tex]
