Question

The common roots of the equations \(x^3+2x^2+2x+1=0\) and \(1+x^{2002}+x^{2003}=0\) are (where \(\omega\) is a complex cube root of unity)

• A. \(\omega,\omega^2\)
• B. \(1,\omega^2\)
• C. \(-1,-\omega\)
• D. \(\omega,-\omega^2\)

Hint:

For cube roots of unity: \[ \omega^3=1,\qquad 1+\omega+\omega^2=0 \] Reduce large powers modulo \(3\).

Answer 1

The Correct Option is A

Concept:
The complex cube roots of unity are: \[ 1,\omega,\omega^2 \] They satisfy: \[ \omega^3=1 \] and: \[ 1+\omega+\omega^2=0 \]

Step 1: Factorize the first equation.
Given: \[ x^3+2x^2+2x+1=0 \] Group terms: \[ x^3+x^2+x^2+x+x+1 \] It can be factorized as: \[ (x+1)(x^2+x+1)=0 \] So roots are: \[ x=-1 \] and roots of: \[ x^2+x+1=0 \] These are: \[ \omega,\omega^2 \] So first equation has roots: \[ -1,\omega,\omega^2 \]

Step 2: Check the second equation.
Second equation: \[ 1+x^{2002}+x^{2003}=0 \] We need to test roots from first equation.

Step 3: Test \(x=\omega\).
Since: \[ \omega^3=1 \] Reduce powers modulo 3. \[ 2002 \equiv 1 \pmod 3 \] because: \[ 2001 \text{ is divisible by }3 \] So: \[ \omega^{2002}=\omega \] Also: \[ 2003 \equiv 2 \pmod 3 \] So: \[ \omega^{2003}=\omega^2 \] Therefore: \[ 1+\omega^{2002}+\omega^{2003} =1+\omega+\omega^2=0 \] So \(\omega\) is a root.

Step 4: Test \(x=\omega^2\).
Using the same property: \[ (\omega^2)^{2002}+(\omega^2)^{2003} \] The powers again reduce to give: \[ 1+\omega^2+\omega=0 \] So \(\omega^2\) is also a root.

Step 5: Test \(x=-1\).
\[ 1+(-1)^{2002}+(-1)^{2003} \] Since: \[ (-1)^{2002}=1 \] and: \[ (-1)^{2003}=-1 \] So: \[ 1+1-1=1\ne0 \] Thus, \(-1\) is not a root of the second equation. Hence, common roots are: \[ \boxed{(A)\ \omega,\omega^2} \]