# How do you factor completely: x^8-9?

Jul 19, 2015

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

#### Explanation:

Using the difference of squares factorisation (${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$) you have:
${x}^{8} - 9 = \left({x}^{4} - 3\right) \left({x}^{4} + 3\right)$

This is probably all they want but you can factor further allowing complex numbers:
$\left({x}^{4} - 3\right) \left({x}^{4} + 3\right) =$
$\left({x}^{2} - {3}^{\frac{1}{2}}\right) \left({x}^{2} + {3}^{\frac{1}{2}}\right) \left({x}^{2} - i {3}^{\frac{1}{2}}\right) \left({x}^{2} + i {3}^{\frac{1}{2}}\right) =$
$\left(x - {3}^{\frac{1}{4}}\right) \left(x + {3}^{\frac{1}{4}}\right) \left(x - i {3}^{\frac{1}{4}}\right) \left(x + i {3}^{\frac{1}{4}}\right) \left(x - \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) {3}^{\frac{1}{4}}\right) \left(x + \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) {3}^{\frac{1}{4}}\right) \left(x - \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) {3}^{\frac{1}{4}}\right) \left(x + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) {3}^{\frac{1}{4}}\right)$

The 8 roots are the 8 solutions to: ${x}^{8} = 9$

Jul 20, 2015

Factor ${x}^{8} - 9$

#### Explanation:

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

= $\left({x}^{2} - \sqrt{3}\right) \left({x}^{2} + \sqrt{3}\right) \left({x}^{4} + 3\right)$

= $\left(x - \sqrt[4]{3}\right) \left(x + \sqrt[4]{3}\right) \left({x}^{2} + \sqrt{3}\right) \left({x}^{4} + 3\right)$