Multiple choice

Let $\alpha$,$\alpha^{2}$ be the roots of $x^{2}+x+1=0$, then the equation whose roots are $\alpha^{31}$ and $\alpha^{62}$ is

  1. $x^{2}-x+1=0$
  2. $x^{2}+x-1=0$
  3. $x^{2}+x+1=0$
  4. $2x^{2}+x+1=0$
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C Correct answer
Explanation

Roots of x^2+x+1=0 are omega and omega^2. Alpha = omega, so alpha^31 = omega^31 = omega. Alpha^62 = omega^62 = omega^2. The equation with roots omega and omega^2 is x^2+x+1=0.

AI explanation

Given that alpha and alpha^2 are roots of x^2 + x + 1 = 0, we know from the properties of roots of unity that alpha^3 = 1. This means any power of alpha can be reduced by taking the exponent modulo 3. For the new roots, alpha^31 is equivalent to alpha^(3*10 + 1), which equals alpha, and alpha^62 is equivalent to alpha^(3*20 + 2), which equals alpha^2. Since the new roots are identical to the original roots, the required equation remains x^2 + x + 1 = 0.

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