Multiple choice

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

  1. $ { y }^{ 3 }-r{ y }^{ 2 }+\left( 4r-{ q }^{ 2 } \right) { y }+{ p\left( 8r-4pq+{ p }^{ 3 } \right) }=0$
  2. $ { y }^{ 3 }-p{ y }^{ 2 }+\left( 4q-{ p }^{ 2 } \right) { y }+{ p\left( 8r-4pq+{ p }^{ 3 } \right) }=0$
  3. $ { ry }^{ 3 }-q\left( q+1 \right) { y }^{ 2 }+{ p\left( r+1 \right) }^{ 2 }y-{ \left( p+1 \right) }^{ 3 }=0$
  4. $ { ry }^{ 3 }-q\left( r+1 \right) { y }^{ 2 }+{ p\left( r+1 \right) }^{ 2 }y-{ \left( r+1 \right) }^{ 3 }=0$
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B Correct answer
AI explanation

Using Vieta's formulas for x^3 - px^2 + qx - r = 0, we have the sum of roots alpha + beta + gamma = p. The new roots can be rewritten using this sum as (p - 2*alpha), (p - 2*beta), and (p - 2*gamma). Applying the linear transformation y = p - 2x to the original polynomial gives x = (p - y)/2, which we substitute into the cubic equation: ((p - y)/2)^3 - p((p - y)/2)^2 + q((p - y)/2) - r = 0. Expanding this and multiplying through by -8 to make it a monic polynomial in terms of y results in the equation y^3 - py^2 + (4q - p^2)y + p(8r - 4pq + p^3) = 0.

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