How do you factor $27 a^{3} - 8$ $27a^3-8$ ?

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Answer 1 · 2018-04-07T18:02:47

$27 a^{3} - 8 = (3 a - 2) (9 a^{2} + 6 a + 4)$ $27a^3-8=color(red)((3a-2)(9a^2+6a+4))$

We have a difference of cubes:

$27 a^{3} - 8$ $27a^3-8$

$= {(3 a)}^{3} - 2^{3}$ $=(3a)^3-2^3$

To factor a difference of cubes, use the formula:

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

${(3 a)}^{3} - 2^{3} = (3 a - 2) (9 a^{2} + 6 a + 4)$ $(3a)^3-2^3=color(red)((3a-2)(9a^2+6a+4))$

This cannot be factored any further using rational numbers, so this is the fully factored form.

How do you factor 27a3−827a^3-8?