[tex]1\text{ and }\frac{\text{-1 }}{2}\pm\text{ }\frac{i\sqrt[]{3^{}}}{2}\text{ (option C)}[/tex]
Explanation:[tex]\begin{gathered} x^3-\text{ 1 = 0} \\ x^3-\text{ 1 has a root of 1} \\ x^3-1=(x-1)(x^2\text{ + x + 1)} \end{gathered}[/tex]
we find the root of x² + x + 1 has it can't be factorized
Using quadratic formula:
[tex]x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
for a² + bx + c = 0
comparing: x² + x + 1
where a = 1, b = 1, c = 1
[tex]\begin{gathered} x\text{ = }\frac{-1\pm\sqrt[]{(1)^2^{}-4(1)(1)}}{2(1)} \\ x\text{ = }\frac{-1\pm\sqrt[]{1^{}-4}}{2} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = }\frac{-1\pm\sqrt[]{-3}}{2}\text{= }\frac{-1\pm\sqrt[]{-1(3)}}{2} \\ Since\text{ we can't find the square root of a negative number, we apply complex root} \\ \text{let i}^2\text{ = -1} \\ x\text{ = }\frac{-1\pm\sqrt[]{3i^2}}{2} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = }\frac{-1\pm\sqrt[]{3i^2}}{2}\text{ = }\frac{-1\pm i\sqrt[]{3^{}}}{2} \\ x\text{ = }\frac{-1+i\sqrt[]{3^{}}}{2}or\text{ }\frac{-1-i\sqrt[]{3^{}}}{2} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{The roots of x}^3\text{ - 1 = 0 are:} \\ 1\text{ and }\frac{-1\pm i\sqrt[]{3^{}}}{2} \\ 1\text{ and }\frac{\text{-1 }}{2}\pm\text{ }\frac{i\sqrt[]{3^{}}}{2}\text{ (option C)} \end{gathered}[/tex]
