Factor completely. x^6-y^6

1 answer

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 \( x^3 + y^3 \).

  1. 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, \[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]

  2. 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:

\[ (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2) \]