# Factorizing Problem

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• May 16th 2012, 12:59 AM
BobBali
Factorizing Problem
Hello All,

Need a little help finding the roots for:

$\displaystyle m^6 - 8$
Let $\displaystyle A= x^2$
$\displaystyle (a)^3 - 8= 0$

$\displaystyle (a)^3 = 8 Therefore, a^3 = 8 so, a^2 = 8 hence, (x^2 - 2) is one root$

$\displaystyle (x^2 - 2) (X^4 + bx + 4) = 0$

After this how do i find the 'b' coefficient...?
• May 16th 2012, 04:03 PM
Reckoner
Re: Factorizing Problem
Quote:

Originally Posted by BobBali
$\displaystyle m^6 - 8$

This is a difference of cubes:

$\displaystyle m^6-8 = \left(m^2\right)^3 - 2^3 = \left(m^2-2\right)\left(m^4+2m^2+4\right)$

Set each factor equal to zero. The second factor is actually quadratic in form, so we can use the quadratic formula to find its roots.

$\displaystyle m^6-8=\left(m^2-2\right)\left[\left(m^2\right)^2 + 2\left(m^2\right)+4\right]$

So

$\displaystyle m^2-2 = 0\Rightarrow m=\pm\sqrt2$

and

$\displaystyle \left(m^2\right)^2 + 2\left(m^2\right)+4=0\Rightarrow m^2 = \frac{-2\pm\sqrt{-12}}2$

$\displaystyle \Rightarrow m^2 = -1\pm i\sqrt3$

$\displaystyle \Rightarrow m = \pm\sqrt{-1+i\sqrt3}\mathrm{\quad or\quad}m=\pm\sqrt{-1-i\sqrt3}$