Factor polynomial $$ d^{8}-1 $$

$$ d^{8}-1 = ( d-1 )( d+1 )( d^{2}+1 )( d^{4}+1 ) $$

Step 1 :

Rewrite $ d^{8}-1 $ as:

$$ d^{8}-1 = (d^{4})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = d^{4} $ and $ II = 1 $ , we have:

$$ d^{8}-1 = (d^{4})^2 - (1)^2 = ( d^{4}-1 ) ( d^{4}+1 ) $$

Step 2 :

Rewrite $ d^{4}-1 $ as:

$$ d^{4}-1 = (d^{2})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = d^{2} $ and $ II = 1 $ , we have:

$$ d^{4}-1 = (d^{2})^2 - (1)^2 = ( d^{2}-1 ) ( d^{2}+1 ) $$

Step 3 :

Rewrite $ d^{2}-1 $ as:

$$ d^{2}-1 = (d)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = d $ and $ II = 1 $ , we have:

$$ d^{2}-1 = (d)^2 - (1)^2 = ( d-1 ) ( d+1 ) $$

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