Step 1 :

Rewrite $ x^{12}-1 $ as:

$$ x^{12}-1 = (x^{6})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = x^{6} $ and $ II = 1 $ , we have:

$$ x^{12}-1 = (x^{6})^2 - (1)^2 = ( x^{6}-1 ) ( x^{6}+1 ) $$

Step 2 :

Rewrite $ x^{6}-1 $ as:

$$ x^{6}-1 = (x^{3})^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = x^{3} $ and $ II = 1 $ , we have:

$$ x^{6}-1 = (x^{3})^2 - (1)^2 = ( x^{3}-1 ) ( x^{3}+1 ) $$

Step 3 :

To factor $ x^{3}-1 $ we can use difference of cubes formula:

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

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

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

Step 4 :

Step 4: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.

Step 5: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 5 :

To factor $ x^{3}+1 $ we can use sum of cubes formula:

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

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

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

Step 6 :

Step 6: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.

Step 7: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 7 :

Step 7: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ 1 } $.

Step 8: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 9: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.

Step 8 :

To factor $ x^{6}+1 $ we can use sum of cubes formula:

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

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

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

Step 9 :

Step 9: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.

Step 10: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 11: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

Step 10 :

Step 10: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ 1 } $.

Step 11: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 12: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.

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