Multiple choice

If $\alpha ,\beta ,\gamma $ are the roots of the cubic ${ x }^{ 3 }+qx+r=0$, then the equation whose roots are $ { \left( \alpha -\beta \right) }^{ 2 },{ \left( \beta -\gamma \right) }^{ 2 },{ \left( \gamma -\alpha \right) }^{ 2 }$, is:

  1. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4q }^{ 3 }+{ 27r }^{ 2 } \right) =0$
  2. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4r }^{ 3 }-{ 27r }^{ 2 } \right) =0$
  3. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4r }^{ 3 }+{ 27q }^{ 2 } \right) =0$
  4. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4q }^{ 3 }-{ 27r }^{ 2 } \right) =0$
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A Correct answer
Explanation

This is a standard problem in the theory of equations. For the cubic x^3 + qx + r = 0, the transformation to find the equation with roots (alpha-beta)^2 etc., leads to the result y^3 + 6qy^2 + 9q^2y + (4q^3 + 27r^2) = 0.

AI explanation

For the cubic equation x^3 + qx + r = 0, the sum of the roots alpha + beta + gamma = 0. The discriminant of the cubic is given by the formula -(4q^3 + 27r^2). To find the equation whose roots are the squared differences of the original roots, we know the product of these squared differences equals the discriminant. Therefore, the constant term of the new cubic equation (the product of its roots) must be 4q^3 + 27r^2. Matching this with the required sum and product of the squared differences yields the new equation y^3 + 6q y^2 + 9q^2 y + (4q^3 + 27r^2) = 0.

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