Factor polynomial $$ x^{100}-1 $$

$$ x^{100}-1 = ( x-1 )( x^{4}+x^{3}+x^{2}+x+1 )( x^{20}+x^{15}+x^{10}+x^{5}+1 )( x+1 )( x^{4}-x^{3}+x^{2}-x+1 )( x^{20}-x^{15}+x^{10}-x^{5}+1 )( x^{2}+1 )( x^{8}-x^{6}+x^{4}-x^{2}+1 )( x^{40}-x^{30}+x^{20}-x^{10}+1 ) $$

Step 1 :

Rewrite $ x^{100}-1 $ as:

$$ x^{100}-1 = (x^{50})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = x^{50} $ and $ II = 1 $ , we have:

$$ x^{100}-1 = (x^{50})^2 - (1)^2 = ( x^{50}-1 ) ( x^{50}+1 ) $$

Step 2 :

Rewrite $ x^{50}-1 $ as:

$$ x^{50}-1 = (x^{25})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = x^{25} $ and $ II = 1 $ , we have:

$$ x^{50}-1 = (x^{25})^2 - (1)^2 = ( x^{25}-1 ) ( x^{25}+1 ) $$

Step 3 :

To factor $ x^{25}-1 $ we can use formula:

$$ I^5 - II^5 = (I - II)(I^4 + I^3 \cdot II + I^2 \cdot II^2 + I \cdot II^3 + II^4) $$

After putting $ I = x^{5} $ and $ II = 1 $ , we have:

$$ x^{25}-1 = ( x^{5}-1 ) ( x^{20}+x^{15}+x^{10}+x^{5}+1 ) $$

Step 4 :

To factor $ x^{5}-1 $ we can use formula:

$$ I^5 - II^5 = (I - II)(I^4 + I^3 \cdot II + I^2 \cdot II^2 + I \cdot II^3 + II^4) $$

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

$$ x^{5}-1 = ( x-1 ) ( x^{4}+x^{3}+x^{2}+x+1 ) $$

Step 5 :

To factor $ x^{25}+1 $ we can use formula:

$$ I^5 + II^5 = (I + II)(I^4 - I^3 \cdot II + I^2 \cdot II^2 - I \cdot II^3 + II^4) $$

After putting $ I = x^{5} $ and $ II = 1 $ , we have:

$$ x^{25}+1 = ( x^{5}+1 ) ( x^{20}-x^{15}+x^{10}-x^{5}+1 ) $$

Step 6 :

To factor $ x^{5}+1 $ we can use formula:

$$ I^5 + II^5 = (I + II)(I^4 - I^3 \cdot II + I^2 \cdot II^2 - I \cdot II^3 + II^4) $$

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

$$ x^{5}+1 = ( x+1 ) ( x^{4}-x^{3}+x^{2}-x+1 ) $$

Step 7 :

To factor $ x^{50}+1 $ we can use formula:

$$ I^5 + II^5 = (I + II)(I^4 - I^3 \cdot II + I^2 \cdot II^2 - I \cdot II^3 + II^4) $$

After putting $ I = x^{10} $ and $ II = 1 $ , we have:

$$ x^{50}+1 = ( x^{10}+1 ) ( x^{40}-x^{30}+x^{20}-x^{10}+1 ) $$

Step 8 :

To factor $ x^{10}+1 $ we can use formula:

$$ I^5 + II^5 = (I + II)(I^4 - I^3 \cdot II + I^2 \cdot II^2 - I \cdot II^3 + II^4) $$

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

$$ x^{10}+1 = ( x^{2}+1 ) ( x^{8}-x^{6}+x^{4}-x^{2}+1 ) $$

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