Step 1 :

Rewrite $ x^12-y^12 $ as:

$$ \color{blue}{ x^12-y^12 = (x^6)^2 - (y^6)^2 } $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x^6 $ and $ II = y^6 $ , we have:

$$ x^12-y^12 = (x^6)^2 - (y^6)^2 = ( x^6-y^6 ) ( x^6+y^6 ) $$

Step 2 :

To factor $ x^{6}+y^{6} $ we can use sum of cubes formula:

$$ I^3 + II^3 = (I + II) (I^2 - I \cdot II + II^2)$$

After putting $ I = x^2 $ and $ II = y^2 $ , we have:

$$ x^{6}+y^{6} = ( x^{2}+y^{2} ) ( x^{4}-x^{2}y^{2}+y^{4} ) $$

Step 3 :

Rewrite $ x^6-y^6 $ as:

$$ \color{blue}{ x^6-y^6 = (x^3)^2 - (y^3)^2 } $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x^3 $ and $ II = y^3 $ , we have:

$$ x^6-y^6 = (x^3)^2 - (y^3)^2 = ( x^3-y^3 ) ( x^3+y^3 ) $$

Step 4 :

To factor $ x^{3}+y^{3} $ we can use sum of cubes formula:

$$ I^3 + II^3 = (I + II) (I^2 - I \cdot II + II^2)$$

After putting $ I = x $ and $ II = y $ , we have:

$$ x^{3}+y^{3} = ( x+y ) ( x^{2}-xy+y^{2} ) $$

Step 5 :

To factor $ x^{3}-y^{3} $ we can use difference of cubes formula:

$$ I^3 - II^3 = (I - II) (I^2 + I \cdot II + II^2) $$

After putting $ I = x $ and $ II = y $ , we have:

$$ x^{3}-y^{3} = ( x-y ) ( x^{2}+xy+y^{2} ) $$

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