Step 1 :

To factor $ x^{4}-x^{3}-27x+27 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{4} }$ with $ \color{blue}{ -x^{3} }$ and $ \color{red}{ -27x }$ with $ \color{red}{ 27 }$ then factor each group.

$$ \begin{aligned} x^{4}-x^{3}-27x+27 = ( \color{blue}{ x^{4}-x^{3} } ) + ( \color{red}{ -27x+27 }) &= \\ &= \color{blue}{ x^{3}( x-1 )} + \color{red}{ -27( x-1 ) } = \\ &= (x^{3}-27)(x-1) \end{aligned} $$

Step 2 :

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

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

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:

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

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

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 9 }$.

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

Step 5: 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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