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Semmy [17]
2 years ago
6

Solve the equation in the complex number system.

Mathematics
2 answers:
Anna35 [415]2 years ago
6 0
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} \}
tatiyna2 years ago
3 0
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.
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