Question

If a complex number \(\alpha\) is a common root of \[ x^{2026}+x^{1964}+1=0 \] and \[ x^3+2x^2+2x+1=0, \] then the sum of the complex roots of the equation \[ z^3=\alpha^3 \] is

• A. \(2+3i\)
• B. \(-\dfrac12+\dfrac{\sqrt3}{2}i\)
• C. \(1\)
• D. \(-1\)

Hint:

The cube roots of unity are \[ 1,\quad \omega,\quad \omega^2, \] where \[ \boxed{1+\omega+\omega^2=0.} \] Hence, the sum of the two non-real cube roots is \[ \boxed{\omega+\omega^2=-1.} \]

Answer 1

The Correct Option is D

Step 1: Find the common root \(\alpha\). Factor the cubic polynomial: \[ x^3+2x^2+2x+1 = (x+1)(x^2+x+1). \] Hence the roots are \[ -1,\qquad \frac{-1\pm i\sqrt3}{2}. \] Now, \[ 2026\equiv1\pmod3,\qquad 1964\equiv2\pmod3. \] For a cube root of unity, \[ \omega^3=1, \] so \[ \omega^{2026}=\omega,\qquad \omega^{1964}=\omega^2. \] Therefore, \[ \omega^{2026}+\omega^{1964}+1 = \omega+\omega^2+1 =0. \] Thus, the common root is \[ \boxed{\alpha=\omega \text{ or } \omega^2.} \]

Step 2: Form the equation \(z^3=\alpha^3\). Since \[ \alpha^3=1, \] the equation becomes \[ z^3=1. \] Its three roots are \[ 1,\qquad \omega,\qquad \omega^2. \]

Step 3: Find the sum of the roots. The sum of the cube roots of unity is \[ 1+\omega+\omega^2=0. \] However, using \[ x^3-1=0, \] the sum of the complex (non-real) roots is \[ \omega+\omega^2 = -1. \] Hence, \[ \boxed{-1} \] is the correct answer. Thus, \[ \boxed{(D)} \] is the correct answer.