Answer to Question #251784 in Linear Algebra for Qzl.q

Question

Find all the third root of( -8)?
(Complex numbers)

Answer 1

Let X be the cube root of -8.

$X^3 = -8$

$X^3 + 8=0$

$(X+2)(X^2-2X+4)=0$

For X + 2 = 0; we have X = −2

$X^2-2X+4=0$

This is need to solve this using the quadratic formula:

$X=\frac{-(-2)\pm\sqrt{(-2)^2-4\times1\times4}}{2\times1}$

$X=\frac{2\pm\sqrt{4-16}}{2}$

$X=\frac{2\pm\sqrt{-12}}{2}$

$X=\frac{2\pm2\sqrt{-3}}{2}$

$X=1\pm\sqrt{-3}$

This means all the third root of( -8) are;

$-2, 1+i\sqrt{3}, 1-i\sqrt{3}$