Which of the following expressions are equivalent to 2/x^8-y^8? choose all that apply
2 answers:
2/x^8-y^8 =2/(x^4-y^4)^2=2/(x^4-y^4) * 1/(x^4-y^4)
Answer:
Option A and B are correct
and ![\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E4-y%5E4%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7Bx%5E4%2By%5E4%7D)
Step-by-step explanation:
Using the identity rule:
![x^2-y^2 = (x+y)(x-y)](https://tex.z-dn.net/?f=x%5E2-y%5E2%20%3D%20%28x%2By%29%28x-y%29)
Given the expression:
![\frac{2}{x^8-y^8}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E8-y%5E8%7D)
We can write this as:
![\frac{2}{(x^4)^2-(y^4)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28x%5E4%29%5E2-%28y%5E4%29%5E2%7D)
Apply the identity rules:
![\frac{2}{(x^4-y^4)(x^4+y^4)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28x%5E4-y%5E4%29%28x%5E4%2By%5E4%29%7D)
We can write this as:
![\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E4-y%5E4%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7Bx%5E4%2By%5E4%7D)
Therefore, the expressions which is equivalent to
are:
and ![\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E4-y%5E4%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7Bx%5E4%2By%5E4%7D)
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