Step 1 :
To factor $ w^{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 = w $ and $ II = 1 $ , we have:
$$ w^{3}-1 = ( w-1 ) ( w^{2}+w+1 ) $$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:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
PRODUCT = 1 | |
1 1 | -1 -1 |
Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.