Step 1 :

To factor $ 512y^{3}+1 $ we can use sum of cubes formula:

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

After putting $ I = 8y $ and $ II = 1 $ , we have:

$$ 512y^{3}+1 = ( 8y+1 ) ( 64y^{2}-8y+1 ) $$

Step 2 :

Step 2: 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:

$$ a = 64 , b = -8 ~ \text{ and } ~ c = 1 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 64 }$ by the constant term $\color{blue}{c = 1} $.

$$ a \cdot c = 64 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 64 $ and add to $ b = -8 $.

Step 5: All pairs of numbers with a product of $ 64 $ are:

PRODUCT = 64
1     64	-1     -64
2     32	-2     -32
4     16	-4     -16
8     8	-8     -8

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -8 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -8 }$, we conclude the polynomial cannot be factored.

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