Multiple choice

If $a,b,c$ are the roots of $x^3+8=0$, then the equation whose roots are $a^2,b^2,c^2$ is-

  1. $x^3-8=0$
  2. $x^3+64=0$
  3. $x^3+16=0$
  4. $x^3-64=0$
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D Correct answer
Explanation

If x^3 = -8, then the roots a, b, c satisfy a^3 = -8, b^3 = -8, c^3 = -8. We want an equation with roots a^2, b^2, c^2. Let y = x^2, so x = y^(1/2). Substituting into x^3 = -8 gives (y^(1/2))^3 = -8, so y^(3/2) = -8. Squaring both sides gives y^3 = 64, or y^3 - 64 = 0.

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