Factor polynomial $$ 64v^{6}-y^{6} $$

$$ 64v^{6}-y^{6} = (2v+y)(4v^{2}-2vy+y^{2})(2v-y)(4v^{2}+2vy+y^{2}) $$

Step 1 :

Rewrite $ 64v^6-y^6 $ as:

$$ \color{blue}{ 64v^6-y^6 = (8v^3)^2 - (y^3)^2 } $$

Now we can apply the difference of squares formula.

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

After putting $ I = 8v^3 $ and $ II = y^3 $ , we have:

$$ 64v^6-y^6 = (8v^3)^2 - (y^3)^2 = ( 8v^3-y^3 ) ( 8v^3+y^3 ) $$

Step 2 :

To factor $ 8v^{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 = 2v $ and $ II = y $ , we have:

$$ 8v^{3}+y^{3} = ( 2v+y ) ( 4v^{2}-2vy+y^{2} ) $$

Step 3 :

To factor $ 8v^{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 = 2v $ and $ II = y $ , we have:

$$ 8v^{3}-y^{3} = ( 2v-y ) ( 4v^{2}+2vy+y^{2} ) $$

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