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Sagot :
To determine which expressions are equivalent to [tex]\(\frac{2}{x^6 - y^6}\)[/tex], let's analyze and simplify each expression.
First, recall that [tex]\(x^6 - y^6\)[/tex] can be factored using the difference of powers formula:
[tex]\[ x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \][/tex]
So we can rewrite:
[tex]\[ \frac{2}{x^6 - y^6} = \frac{2}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
### Option A
For option A:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} = \frac{2 \cdot 1}{(x^3 - y^3)(x^3 + y^3)} = \frac{2}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
This matches [tex]\(\frac{2}{x^6 - y^6}\)[/tex].
### Option B
For option B:
[tex]\[ \frac{1}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{1}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} = \frac{1 \cdot 1}{(x^3 - y^3)(x^3 + y^3)} = \frac{1}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
This does not match [tex]\(\frac{2}{x^6 - y^6}\)[/tex], as the numerator is [tex]\(1\)[/tex] instead of [tex]\(2\)[/tex].
### Option C
For option C:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 - y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 - y^3} = \frac{2 \cdot 1}{(x^3 - y^3)^2} = \frac{2}{(x^3 - y^3)^2} \][/tex]
This does not match [tex]\(\frac{2}{x^6 - y^6}\)[/tex], as the denominator is [tex]\((x^3 - y^3)^2\)[/tex] instead of [tex]\((x^3 - y^3)(x^3 + y^3)\)[/tex].
### Option D
For option D:
[tex]\[ \frac{2}{(x^3)^2 - (y^3)^2} \][/tex]
Recognizing the difference of squares:
[tex]\[ (x^3)^2 - (y^3)^2 = x^6 - y^6 \][/tex]
So:
[tex]\[ \frac{2}{(x^3)^2 - (y^3)^2} = \frac{2}{x^6 - y^6} \][/tex]
This matches [tex]\(\frac{2}{x^6 - y^6}\)[/tex].
### Conclusion
The expressions that are equivalent to [tex]\(\frac{2}{x^6 - y^6}\)[/tex] are:
- Option A: [tex]\(\frac{2}{\left(x^3-y^3\right)} \cdot \frac{1}{\left(x^3+y^3\right)}\)[/tex]
- Option D: [tex]\(\frac{2}{\left(x^3\right)^2-\left(y^3\right)^2}\)[/tex]
Therefore, the correct options are [tex]\(\boxed{\text{A and D}}\)[/tex].
First, recall that [tex]\(x^6 - y^6\)[/tex] can be factored using the difference of powers formula:
[tex]\[ x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \][/tex]
So we can rewrite:
[tex]\[ \frac{2}{x^6 - y^6} = \frac{2}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
### Option A
For option A:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} = \frac{2 \cdot 1}{(x^3 - y^3)(x^3 + y^3)} = \frac{2}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
This matches [tex]\(\frac{2}{x^6 - y^6}\)[/tex].
### Option B
For option B:
[tex]\[ \frac{1}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{1}{x^3 - y^3} \cdot \frac{1}{x^3 + y^3} = \frac{1 \cdot 1}{(x^3 - y^3)(x^3 + y^3)} = \frac{1}{(x^3 - y^3)(x^3 + y^3)} \][/tex]
This does not match [tex]\(\frac{2}{x^6 - y^6}\)[/tex], as the numerator is [tex]\(1\)[/tex] instead of [tex]\(2\)[/tex].
### Option C
For option C:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 - y^3} \][/tex]
Combining the fractions:
[tex]\[ \frac{2}{x^3 - y^3} \cdot \frac{1}{x^3 - y^3} = \frac{2 \cdot 1}{(x^3 - y^3)^2} = \frac{2}{(x^3 - y^3)^2} \][/tex]
This does not match [tex]\(\frac{2}{x^6 - y^6}\)[/tex], as the denominator is [tex]\((x^3 - y^3)^2\)[/tex] instead of [tex]\((x^3 - y^3)(x^3 + y^3)\)[/tex].
### Option D
For option D:
[tex]\[ \frac{2}{(x^3)^2 - (y^3)^2} \][/tex]
Recognizing the difference of squares:
[tex]\[ (x^3)^2 - (y^3)^2 = x^6 - y^6 \][/tex]
So:
[tex]\[ \frac{2}{(x^3)^2 - (y^3)^2} = \frac{2}{x^6 - y^6} \][/tex]
This matches [tex]\(\frac{2}{x^6 - y^6}\)[/tex].
### Conclusion
The expressions that are equivalent to [tex]\(\frac{2}{x^6 - y^6}\)[/tex] are:
- Option A: [tex]\(\frac{2}{\left(x^3-y^3\right)} \cdot \frac{1}{\left(x^3+y^3\right)}\)[/tex]
- Option D: [tex]\(\frac{2}{\left(x^3\right)^2-\left(y^3\right)^2}\)[/tex]
Therefore, the correct options are [tex]\(\boxed{\text{A and D}}\)[/tex].
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