Question

Solve the equation in the complex number system.

X^-3 - 1=0.

I know you have to move the 1 over which leaves you with

X^3=1. How do you get rid of the x^3?

Answer 1

`x^3-1=0\\ (x-1)(x^2+x+1)=0\\ x-1=0\\ x=1\\ x^2+x+1=0\\ x^2+x+(1)/(4)+(3)/(4)=0\\ (x+(1)/(2))^2=-(3)/(4)\\ x+(1)/(2)=\sqrt{-(3)/(4)} \vee x+(1)/(2)=-\sqrt{-(3)/(4)}\\ x=-(1)/(2)+i(\sqrt3)/(2) \vee x=-(1)/(2)-i(\sqrt3)/(2)\\ x=-(1-i\sqrt3)/(2) \vee x=-(1+i\sqrt3)/(2)\\\\ x=\{1,-(1-i\sqrt3)/(2),-(1+i\sqrt3)/(2) \}`

Answer 2

To get rid of
`x^(3)`, you have to take the third root of both sides:

`\sqrt[3]{x^(3)} = \sqrt[3]{1}`
But that won't help you with understanding the problem. It is better to write
`x^(3)-1` as a product of 2 polynomials:

`x^(3)-1 = (x-1)\cdot (x^(2) +x +1)`
From this we know, that
`x-1 = 0 => x = 1` is the solution. Another solutions (complex roots) are the roots of quadratic equation.