Answered

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?

Verified Answer :

mashenka

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.

konrad509

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

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