The common roots of the equation $Z^3+2z^2+2z+1=0, z^{223}+z^{224}+1=0$ are
- $\omega,\omega^2$
- $1,\omega,\omega^2$
- $-1,\omega.\omega^2$
- $-\omega,-\omega^2$
Reveal answer
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A
Correct answer
Explanation
z^3+2z^2+2z+1 = (z+1)(z^2+z+1) = 0. Roots are -1, omega, omega^2. For z^223+z^224+1=0: If z=-1, (-1)^223+(-1)^224+1 = -1+1+1 = 1 (not 0). If z=omega, omega^223+omega^224+1 = omega^1+omega^2+1 = 0. If z=omega^2, omega^446+omega^448+1 = omega^2+omega^1+1 = 0.
AI explanation
The first equation factors as (z+1)(z^2+z+1)=0, yielding roots of -1, omega, and omega^2. For the second equation, z cannot be -1 because (-1)^223 + (-1)^224 + 1 equals 1 instead of 0. However, substituting z = omega gives omega^223 + omega^224 + 1 = omega^2 + 1 + 1, which evaluates to 0 when using omega^3 = 1, and similarly for omega^2. Thus, the common roots are omega and omega^2.