Step 1 :
To factor $ 216x^{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 = 6x $ and $ II = 1 $ , we have:
$$ 216x^{3}+1 = ( 6x+1 ) ( 36x^{2}-6x+1 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 36 }$ by the constant term $\color{blue}{c = 1} $.
$$ a \cdot c = 36 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 36 $ and add to $ b = -6 $.
Step 5: All pairs of numbers with a product of $ 36 $ are:
PRODUCT = 36 | |
1 36 | -1 -36 |
2 18 | -2 -18 |
3 12 | -3 -12 |
4 9 | -4 -9 |
6 6 | -6 -6 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -6 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -6 }$, we conclude the polynomial cannot be factored.