Factor polynomial $$ 8x^{3}-729 $$

$$ 8x^{3}-729 = ( 2x-9 )( 4x^{2}+18x+81 ) $$

Step 1 :

To factor $ 8x^{3}-729 $ we can use difference of cubes formula:

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

After putting $ I = 2x $ and $ II = 9 $ , we have:

$$ 8x^{3}-729 = ( 2x-9 ) ( 4x^{2}+18x+81 ) $$

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:

$$ a = 4 , b = 18 ~ \text{ and } ~ c = 81 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 81} $.

$$ a \cdot c = 324 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 324 $ and add to $ b = 18 $.

Step 5: All pairs of numbers with a product of $ 324 $ are:

PRODUCT = 324
1     324	-1     -324
2     162	-2     -162
3     108	-3     -108
4     81	-4     -81
6     54	-6     -54
9     36	-9     -36
12     27	-12     -27
18     18	-18     -18

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 18 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 18 }$, we conclude the polynomial cannot be factored.

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