Factor polynomial $$ 3y^{3}-192 $$

$$ 3y^{3}-192 = 3( y-4 )( y^{2}+4y+16 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 3y^{3}-192 = 3 ( y^{3}-64 ) $$

Step 2 :

To factor $ y^{3}-64 $ we can use difference of cubes formula:

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

After putting $ I = y $ and $ II = 4 $ , we have:

$$ y^{3}-64 = ( y-4 ) ( y^{2}+4y+16 ) $$

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:

$$ \color{blue}{ b = 4 } ~ \text{ and } ~ \color{red}{ c = 16 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 4 } $ and multiply to $ \color{red}{ 16 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 16 }$.

PRODUCT = 16
1     16	-1     -16
2     8	-2     -8
4     4	-4     -4

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 4 }$, we conclude the polynomial cannot be factored.

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