Question

which of the following expressions are equivalent to 2/x^8-y^8? choose all that apply
A.2/(x^4-y^4)*1/(x^4+y^4)
B.2/(x^4)^2-(y^4)^2
C.2/(x4-y^4)*1/(x^4-y^4)
D.1/(x^4-y^4)*1/(x^4+y^4)

Answer 1

2/x^8-y^8 =2/(x^4-y^4)^2=2/(x^4-y^4) * 1/(x^4-y^4)

Answer 2

Answer:

Option A and B are correct

$\frac{2}{(x^4)^2-(y^4)^2}$and  $\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}$

Step-by-step explanation:

Using the identity rule:

$x^2-y^2 = (x+y)(x-y)$

Given the expression:

$\frac{2}{x^8-y^8}$

We can write this as:

$\frac{2}{(x^4)^2-(y^4)^2}$

Apply the identity rules:

$\frac{2}{(x^4-y^4)(x^4+y^4)}$

We can write this as:

$\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}$

Therefore, the expressions which is equivalent to $\frac{2}{x^8-y^8}$ are:$\frac{2}{(x^4)^2-(y^4)^2}$ and  $\frac{2}{x^4-y^4} \times \frac{1}{x^4+y^4}$