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

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

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Answer 1 · 2016-01-04T23:25:04

$a^{3} - 8 = (a - 2) (a^{2} + 2 a + 4)$ $a^3-8 = (a-2)(a^2+2a+4)$

$= (a - 2) (a + 1 - \sqrt{3} i) (a + 1 + \sqrt{3} i)$ $=(a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)$

Explanation:

Use the difference of cubes identity, which can be written:

$A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ $A^3-B^3=(A-B)(A^2+AB+B^2)$

Note that both $a^{3}$ $a^3$ and $8 = 2^{3}$ $8=2^3$ are both perfect cubes.

So we find:

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

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

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

$= (a - 2) (a^{2} + 2 a + 4)$ $=(a-2)(a^2+2a+4)$

The remaining quadratic factor has negative discriminant, so no Real zeros and no linear factors with Real coefficients. It is possible to factor it using Complex coefficients:

$a^{2} + 2 a + 4$ $a^2+2a+4$

$= a^{2} + 2 a + 1 + 3$ $=a^2+2a+1+3$

$= {(a + 1)}^{2} + 3$ $=(a+1)^2+3$

$= {(a + 1)}^{2} - {(\sqrt{3} i)}^{2}$ $=(a+1)^2-(sqrt(3)i)^2$

$= (a + 1 - \sqrt{3} i) (a + 1 + \sqrt{3} i)$ $= (a+1-sqrt(3)i)(a+1+sqrt(3)i)$

So:

$a^{8} - 8 = (a - 2) (a + 1 - \sqrt{3} i) (a + 1 + \sqrt{3} i)$ $a^8-8 = (a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)$

Another way to express the full factoring is:

$a^{3} - 8 = (a - 2) (a - 2 ω) (a - 2 ω^{2})$ $a^3-8 = (a-2)(a-2omega)(a-2omega^2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2 i$ is the primitive Complex cube root of $1$ $1$ .

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

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