Question

If $ 1, \omega, \omega^2 $ are the cube roots of unity, then the roots of the equation $ 8z^3 - 12z^2 + 6z - 28 = 0 $ are:

• A. \( 2,\, 3\omega,\, 3\omega^2 + 1 \)
• B. \( 2,\, \frac{3\omega + 1}{2},\, \frac{3\omega^2 + 1}{2} \)
• C. \( 2,\, \frac{1 + 3\omega}{3},\, \frac{1 + 3\omega^2}{3} \)
• D. \( 2,\, \frac{1 - \omega}{2},\, \frac{1 - \omega^2}{2} \)

Hint:

When cube roots of unity are mentioned, try expressing complex roots using \( \omega \) and \( \omega^2 \), especially if the quadratic yields imaginary numbers.

Answer 1

The Correct Option is B

Step 1: Observe the structure of the polynomial.
The given polynomial is: \[ 8z^3 - 12z^2 + 6z - 28 = 0. \] Try rational root theorem. Let's test \( z = 2 \): \[ 8(2)^3 - 12(2)^2 + 6(2) - 28 = 64 - 48 + 12 - 28 = 0. \] So, \( z = 2 \) is a root. 
Step 2: Perform polynomial division or use synthetic division.
Divide the polynomial by \( (z - 2) \), and factor the cubic as: \[ 8z^3 - 12z^2 + 6z - 28 = (z - 2)(8z^2 + 4z + 14). \] 
Step 3: Solve the quadratic \( 8z^2 + 4z + 14 = 0 \).
Using quadratic formula: \[ z = \frac{-4 \pm \sqrt{(4)^2 - 4 \cdot 8 \cdot 14}}{2 \cdot 8} = \frac{-4 \pm \sqrt{16 - 448}}{16} = \frac{-4 \pm \sqrt{-432}}{16}. \] \[ \sqrt{-432} = \sqrt{-1 \cdot 144 \cdot 3} = 12i\sqrt{3}, \] \[ z = \frac{-4 \pm 12i\sqrt{3}}{16} = \frac{-1 \pm 3i\sqrt{3}}{4}. \] Now, convert the complex roots to expressions in terms of \( \omega \), using: \[ \omega = \frac{-1 + i\sqrt{3}}{2},\quad \omega^2 = \frac{-1 - i\sqrt{3}}{2}. \] Multiply \( \omega \) by 3 and add 1: \[ \frac{3\omega + 1}{2} = \frac{3\left(\frac{-1 + i\sqrt{3}}{2}\right) + 1}{2} = \frac{\frac{-3 + 3i\sqrt{3}}{2} + 1}{2} = \frac{-1 + 3i\sqrt{3}}{4}, \] which matches the complex root found above. Similarly for \( \omega^2 \), the expression also matches. 
Hence, the roots are: \[ 2, \quad \frac{3\omega + 1}{2}, \quad \frac{3\omega^2 + 1}{2}. \]