Factor polynomial $$ w^{12}-z^{12} $$

$$ w^{12}-z^{12} = (w^{2}+z^{2})(w^{4}-w^{2}z^{2}+z^{4})(w+z)(w^{2}-wz+z^{2})(w-z)(w^{2}+wz+z^{2}) $$

Step 1 :

Rewrite $ w^12-z^12 $ as:

$$ \color{blue}{ w^12-z^12 = (w^6)^2 - (z^6)^2 } $$

Now we can apply the difference of squares formula.

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

After putting $ I = w^6 $ and $ II = z^6 $ , we have:

$$ w^12-z^12 = (w^6)^2 - (z^6)^2 = ( w^6-z^6 ) ( w^6+z^6 ) $$

Step 2 :

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

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

After putting $ I = w^2 $ and $ II = z^2 $ , we have:

$$ w^{6}+z^{6} = ( w^{2}+z^{2} ) ( w^{4}-w^{2}z^{2}+z^{4} ) $$

Step 3 :

Rewrite $ w^6-z^6 $ as:

$$ \color{blue}{ w^6-z^6 = (w^3)^2 - (z^3)^2 } $$

Now we can apply the difference of squares formula.

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

After putting $ I = w^3 $ and $ II = z^3 $ , we have:

$$ w^6-z^6 = (w^3)^2 - (z^3)^2 = ( w^3-z^3 ) ( w^3+z^3 ) $$

Step 4 :

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

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

After putting $ I = w $ and $ II = z $ , we have:

$$ w^{3}+z^{3} = ( w+z ) ( w^{2}-wz+z^{2} ) $$

Step 5 :

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

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

After putting $ I = w $ and $ II = z $ , we have:

$$ w^{3}-z^{3} = ( w-z ) ( w^{2}+wz+z^{2} ) $$

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