Factor polynomial $$ a^{9}+1 $$

$$ a^{9}+1 = ( a+1 )( a^{2}-a+1 )( a^{6}-a^{3}+1 ) $$

Step 1 :

To factor $ a^{9}+1 $ we can use sum of cubes formula:

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

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

$$ a^{9}+1 = ( a^{3}+1 ) ( a^{6}-a^{3}+1 ) $$

Step 2 :

To factor $ a^{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 = a $ and $ II = 1 $ , we have:

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

Step 3 :

Step 3: 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 4: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.

PRODUCT = 1
1     1	-1     -1

Step 5: 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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