Factor polynomial $$ m^{6}-n^{6} $$

$$ m^{6}-n^{6} = (m+n)(m^{2}-mn+n^{2})(m-n)(m^{2}+mn+n^{2}) $$

Step 1 :

Rewrite $ m^6-n^6 $ as:

$$ \color{blue}{ m^6-n^6 = (m^3)^2 - (n^3)^2 } $$

Now we can apply the difference of squares formula.

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

After putting $ I = m^3 $ and $ II = n^3 $ , we have:

$$ m^6-n^6 = (m^3)^2 - (n^3)^2 = ( m^3-n^3 ) ( m^3+n^3 ) $$

Step 2 :

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

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

After putting $ I = m $ and $ II = n $ , we have:

$$ m^{3}+n^{3} = ( m+n ) ( m^{2}-mn+n^{2} ) $$

Step 3 :

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

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

After putting $ I = m $ and $ II = n $ , we have:

$$ m^{3}-n^{3} = ( m-n ) ( m^{2}+mn+n^{2} ) $$

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