# How do you solve x^6 - 9x^3 +8=0?

Jun 12, 2015

This polynomial is quadratic in ${x}^{3}$, factoring into two cubic factors, which then factor a little further...

${x}^{6} - 9 {x}^{3} + 8 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x - 1\right) \left({x}^{2} + x + 1\right)$

So $x = 2$ and $x = 1$ are the real solutions.

#### Explanation:

$0 = {x}^{6} - 9 {x}^{3} + 8$

$= {\left({x}^{3}\right)}^{2} - \left(8 + 1\right) \left({x}^{3}\right) + \left(8 \times 1\right)$

$= \left({x}^{3} - 8\right) \left({x}^{3} - 1\right)$

$= \left({x}^{3} - {2}^{3}\right) \left({x}^{3} - {1}^{3}\right)$

$= \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x - 1\right) \left({x}^{2} + x + 1\right)$

...using the difference of cubes identity (twice):

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Neither of the remaining quadratic factors have linear factors with real coefficients. You can find the discriminant of each to prove that for yourself, or you can factor them using complex numbers as follows:

${x}^{2} + 2 x + 4 = \left(x - 2 \omega\right) \left(x - 2 {\omega}^{2}\right)$

${x}^{2} + x + 1 = \left(x - \omega\right) \left(x - {\omega}^{2}\right)$

Where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

is the primitive complex cube root of unity.

graph{x^6-9x^3+8 [-40.17, 39.83, -18.08, 21.92]}