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

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

Step 1 :

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

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

Now we can apply the difference of squares formula.

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

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

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

Step 2 :

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

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

Now we can apply the difference of squares formula.

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

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

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

Step 3 :

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

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

Now we can apply the difference of squares formula.

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

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

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

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