How do you factor: $y= 27x^3 - 1$ ?

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Answer 1 · 2016-01-18T18:01:18

$(27x^3 - 1) = (3x - 1) (9x^2 + 3x + 1)$

Cube root of 27 is 3 and the cube root of -1 is also -1.
$27x^3 - 1 = (3x)^3 - (1)^3$

To factor $y = 27x^3 - 1$ ,
remember that $(a^3-b^3) = (a-b)(a^2+ab+b^2)$

substituting,

$(27x^3 - 1) = (3x - 1) ((3x)^2 + (3x)(1) + (1)^2)$
$(27x^3 - 1) = (3x - 1) (9x^2 + 3x + 1)$

Answer 2 · 2016-01-18T18:41:05

$(3x-1)(9x^2+3x+1)$

Notice that both the terms are cubed terms, i.e. $27x^3=(3x)^3$ and $1=1^3$ .

That means that this is a difference of cubes, which can be factored as follows;

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Thus, we have $a=3x$ and $b=1$ :

$27x^3-1=(3x)^3-1^3$

$=(3x-1)((3x)^2+(3x)(1)+1^2)$

$=(3x-1)(9x^2+3x+1)$

How do you factor: y= 27x^3 - 1 y= 27x^3 - 1 ?