Q166 / 270 · NDA 2021 (Phase 2) · Mathematics

If $x^2+x+1=0$, then what is the value of $x^{199}+x^{200}+x^{201} ?$

  • A
    -1
  • Correct answer
    B
    0
  • C
    1
  • D
    3

Solution

$\begin{aligned} & \text { Given that } x^2+x+1=0 \\ & \therefore x^{199}+x^{200}+x^{201}=x^{199}+x^{199} \cdot x+x^{199} \cdot x^2 \\ & =x^{199}\left[1+x+x^2\right]=x^{199} \times 0=0\end{aligned}$

Hence, the correct answer is option: 'B'.

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