Answer:
The answer are options
and
![D: (2)/((x^2)^2-(y^2)^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/i2aqe9xcbu7bx7wxi1hdfb1hafly8ntkj1.png)
Explanation:
As all the options are multiplication of fractions the option A cannot be an answer because the numerator multiplication is 1 and different to 2. In the case of option C, observe that if we multiply the denominators we have:
![(x^2-y^2).(x^2-y^2) = (x^2-y^2)^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/16z47xfnmpgqfaoixf5o4vaqq7227pxfr1.png)
As we know for the expanding of the square binomials:
![(x^2-y^2)^2 = (x^2)^2 -2*x^2.y^2 + (y^2)^2 = x^4 +2*x^2y^2 + y^4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ravprtvv016o4xa0wtgz5hrlgqhbt3tdcq.png)
Which is different from the denominator compared:
![x^4 -2*x^2.y^2+y^(4) \\eq x^4 - y^4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2rupdeqovcgkpgrknplbe1z6x8ydlrlpdb.png)
Thus option B cannot be an answer either.
Noting that the denominator compared is a square of two difference by definition, therefore, can be written as:
![x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5truy0o0xe7xz3y4f1v9wfwyi0a5os7s0p.png)
This results in the same denominator as option B. So, option B is a possible answer.
Finally, in the denominator of option D, we can solve the exponents of this factor.
![(x^(2))^(2)-(y^(2))^(2)=x^4 - y^4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8ou6o4axm63u9ggerz8mtvab1u1vctpzae.png)
Which results in the same as the denominator compared, this let option D to be a possible answer.