Help on Solving Equation in the complex number system

[ProdigaI]

Hey, I really need help on this one problem

x^3 - 64 = 0

Thanks

by the way x^3 is x to the 3rd power

 

[Aladdin]

\(\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)\)

\(\displaystyle x^3-64 = (x-4)(x^2+4x+16)\)

\(\displaystyle (x-4)=0\)

\(\displaystyle x=4\)

\(\displaystyle (x^2+4x+16)=0\)

\(\displaystyle Quadratic\)

\(\displaystyle Delta=b'^2-ac = 4-16= -12<0 [In complex numbers -12 is 12(i^2)]\)

\(\displaystyle x=-2+-(i)(sqrt12)/(1\)

[sup:3fo67zfe]Aladdin[/sup:3fo67zfe]

 

[ProdigaI]

thanks : )

 

[soroban]

Hello, ProdigaI!

Are you familiar with the Argand diagram?

\(\displaystyle x^3 - 64 \:=\: 0\)

We have: .\(\displaystyle x^3 \:=\:64 \quadf\Rightarrow\quad x \:=\:4\)

We have a real root: .\(\displaystyle (4,0)\)

On the complex plane, the three cube roots are spaced equally about the origin.
So the graph looks like this:

          *   |
           \  |
            \ |
             \|
      - - - - + - - - - * - -
             /|       (4,0)
            / |
           /  |
          *   |

Using some Trig, we find that the other two points are at:
. . \(\displaystyle \left(\text{-}2,\: 2i\sqrt{3}),\;\;(\text{-}2,\: \text{-} 2i\sqrt{3})\)

Hence, the other two roots are: .\(\displaystyle \text{-}2+2i\sqrt{3}\:\text{ and }\:\text{-}2 -2i\sqrt{3}\)