The expression \( x^6 - y^6 \) is a difference of squares, which can be factored using the identity \( a^2 - b^2 = (a - b)(a + b) \).
In this case:
\[ x^6 - y^6 = (x^3)^2 - (y^3)^2 \]
Applying the difference of squares:
\[ = (x^3 - y^3)(x^3 + y^3) \]
Now, we can further factor \( x^3 - y^3 \) and \( x^3 + y^3 \).
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Factoring \( x^3 - y^3 \): This can be factored using the sum/difference of cubes formula: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] Thus, \[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
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Factoring \( x^3 + y^3 \): This can be factored using the sum of cubes formula: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] Thus, \[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
Putting it all together, we have:
\[ x^6 - y^6 = (x^3 - y^3)(x^3 + y^3) = (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2) \]
So the complete factorization of \( x^6 - y^6 \) is:
\[ (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2) \]