Factor polynomial $$ x^{3}+27 $$

$$ x^{3}+27 = ( x+3 )( x^{2}-3x+9 ) $$

Step 1 :

To factor $ x^{3}+27 $ 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 = 3 $ , we have:

$$ x^{3}+27 = ( x+3 ) ( x^{2}-3x+9 ) $$

Step 2 :

Step 2: 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:

$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = 9 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ 9 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 9 }$.

PRODUCT = 9
1     9	-1     -9
3     3	-3     -3

Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ -3 }$, we conclude the polynomial cannot be factored.

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