Step 1 :

After factoring out $ 9n^{3} $ we have:

$$ 9n^{6}-6561n^{3} = 9n^{3} ( n^{3}-729 ) $$

Step 2 :

To factor $ n^{3}-729 $ we can use difference of cubes formula:

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

After putting $ I = n $ and $ II = 9 $ , we have:

$$ n^{3}-729 = ( n-9 ) ( n^{2}+9n+81 ) $$

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:

$$ \color{blue}{ b = 9 } ~ \text{ and } ~ \color{red}{ c = 81 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 9 } $ and multiply to $ \color{red}{ 81 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 81 }$.

PRODUCT = 81
1     81	-1     -81
3     27	-3     -27
9     9	-9     -9

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 9 }$, we conclude the polynomial cannot be factored.

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