Multiple choice

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

  1. $x^{3}+6x^{2}+9x+1=0$
  2. $x^{3}+6x^{2}+9x-1=0$
  3. $x^{3}+6x^{2}-9x+1=0$
  4. $\left ( x^{2} \right )^{3}-3x^{2}-1=0$
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B Correct answer
Explanation

Let y = x^2, so x = sqrt(y). Substitute into x^3 + 3x - 1 = 0: x(x^2 + 3) = 1. Squaring both sides: x^2(x^2 + 3)^2 = 1. Substituting y: y(y + 3)^2 = 1. y(y^2 + 6y + 9) = 1. y^3 + 6y^2 + 9y - 1 = 0.

AI explanation

For the original cubic x^3 + 3x - 1 = 0, Vieta's formulas give the sum of roots as 0, the sum of products two at a time as 3, and the product of roots as 1. If the new roots are the squares of the original roots, their sum is (sum of roots)^2 - 2*(sum of products) = 0 - 6 = -6. The sum of the products of the new roots taken two at a time is (alpha*beta + beta*gamma + gamma*alpha)^2 - 2*(alpha*beta*gamma)*(alpha + beta + gamma) = 3^2 - 0 = 9. The product of the new roots is (alpha*beta*gamma)^2 = 1. Therefore, the required equation is x^3 - (-6)x^2 + 9x - 1 = 0, which simplifies to x^3 + 6x^2 + 9x - 1 = 0.

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