Question

If \( a \) and \( b \) are the roots of the equation \( y^2 + y + 1 = 0 \), then the value of \( a^4 + b^4 + a^{-1}b^{-1} \) is

• A. \( 1 \)
• B. \( \mathbf{0} \)
• C. \( 5 \)
• D. \( 2 \)

Hint:

Cube roots of unity satisfy \( 1 + \omega + \omega^2 = 0 \). Always check for cyclic identities when such roots appear.

Answer 1

The Correct Option is B

The equation is: \[ y^2 + y + 1 = 0 \Rightarrow y = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{3}i}{2} \] Let \( a = \omega \), \( b = \omega^2 \), where \( \omega \) is a cube root of unity: \[ \omega^3 = 1,\quad 1 + \omega + \omega^2 = 0 \] Now compute: \[ a^4 = \omega^4 = \omega, \quad b^4 = \omega^8 = \omega^2, \quad a^{-1}b^{-1} = (\omega \cdot \omega^2)^{-1} = (\omega^3)^{-1} = 1^{-1} = 1 \] \[ a^4 + b^4 + a^{-1}b^{-1} = \omega + \omega^2 + 1 = 0 \]