Multiple choice

If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+2x+1=0$, then the equation whose roots are $\dfrac{\alpha^{2}}{\beta+\gamma},\dfrac{\beta^{2}}{\gamma +\alpha},\dfrac{\gamma^{2}}{\alpha+\gamma}$ is

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

For the cubic equation x^3 + 2x + 1 = 0, the sum of the roots alpha + beta + gamma = 0, so the denominator beta + gamma equals -alpha. This means the new roots are alpha^2 / (-alpha) = -alpha, -beta, and -gamma. To find the equation whose roots are the negatives of the original roots, substitute (-x) for x in the given equation. This gives (-x)^3 + 2(-x) + 1 = 0, which simplifies to x^3 + 2x - 1 = 0, giving the result x^3 + 2x - 1 = 0.

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