Multiple choice

If $\alpha ,\beta ,\gamma $ are the roots ${ x }^{ 3 }-6x-4=0$, then the equation whose roots are $\displaystyle \beta \gamma +\frac { 1 }{ \alpha } ,\gamma \alpha +\frac { 1 }{ \beta } ,\alpha \beta +\frac { 1 }{ \gamma } $ is

  1. $4{ x }^{ 3 }-30{ x }^{ 2 }+125=0$
  2. ${ x }^{ 3 }+15{ x }^{ 2 }-120=0$
  3. $4{ x }^{ 3 }+30{ x }^{ 2 }-125=0$
  4. $4{ x }^{ 3 }-30{ x }^{ 2 }-125=0$
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AI explanation

From the original equation, Vieta's formulas give the sum of the roots alpha + beta + gamma as 0, the sum of their products as -6, and their product as 4. The new roots are formed by expressions like (beta times gamma) plus (1 divided by alpha), which can be algebraically manipulated to equal 4 divided by alpha squared. Using Newton's sums to find the sum of the inverses of the squares of the roots, we get 15 divided by 4. Multiplying by 4, the sum of the new roots is 15, the sum of their products is -75 divided by 4, and their product is -125 divided by 8. Forming a new cubic polynomial from these sums and clearing the fractions results in 4x^3 - 30x^2 - 125 = 0.

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