Complex Numbers
Cube Roots of Unity
Grade 11
Question:
<p>If the cube roots of unity are \(1, \omega, \omega^2\) then the roots of the equation \((x-1)^3 + 8 = 0\) are</p>
<p>\(-1,\ -1+2\omega,\ -1-2\omega^2\)</p>
<p>\(-1,\ -1,\ -1\)</p>
<p>\(-1,\ 1-2\omega,\ 1-2\omega^2\)</p>
<p>\(-1,\ 1+2\omega,\ 1+2\omega^2\)</p>
Step-by-Step Solution
Key Concept: Recognize that (x-1)³ = -8 means (x-1) equals a cube root of -8. Since cube roots of -8 are -2, -2ω, and -2ω², where ω is a cube root of unity, we get three distinct roots by solving x = 1 + (-2)·(cube root of unity).
<p><strong>Step 1:</strong> Rewrite the equation as $(x-1)^3 = -8$</p><p><strong>Step 2:</strong> The cube roots of -8 are: $-2, -2\omega, -2\omega^2$ where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity.</p><p><strong>Step 3:</strong> Since $(x-1)^3 = -8$, we have $x - 1 \in \{-2, -2\omega, -2\omega^2\}$</p><p><strong>Step 4:</strong> Therefore the three roots are:<br>• $x_1 = 1 + (-2) = -1$<br>• $x_2 = 1 - 2\omega$<br>• $x_3 = 1 - 2\omega^2$</p><p><strong>Verification:</strong> These can also be written as $-1, 1-2\omega, 1-2\omega^2$ where $1 + \omega + \omega^2 = 0$.</p><p>∴ Answer: C</p>
Correct Answer: C