Factor polynomial $$ 128y^{3}-54 $$

$$ 128y^{3}-54 = 2( 4y-3 )( 16y^{2}+12y+9 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 128y^{3}-54 = 2 ( 64y^{3}-27 ) $$

Step 2 :

To factor $ 64y^{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 = 4y $ and $ II = 3 $ , we have:

$$ 64y^{3}-27 = ( 4y-3 ) ( 16y^{2}+12y+9 ) $$

Step 3 :

Step 3: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 16 , b = 12 ~ \text{ and } ~ c = 9 $$

Step 4: Multiply the leading coefficient $\color{blue}{ a = 16 }$ by the constant term $\color{blue}{c = 9} $.

$$ a \cdot c = 144 $$

Step 5: Find out two numbers that multiply to $ a \cdot c = 144 $ and add to $ b = 12 $.

Step 6: All pairs of numbers with a product of $ 144 $ are:

PRODUCT = 144
1     144	-1     -144
2     72	-2     -72
3     48	-3     -48
4     36	-4     -36
6     24	-6     -24
8     18	-8     -18
9     16	-9     -16
12     12	-12     -12

Step 7: Find out which factor pair sums up to $\color{blue}{ b = 12 }$

Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ 12 }$, we conclude the polynomial cannot be factored.

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