Multiple choice

if $\alpha$, $\beta$ are the roots of the equation $ { x }^{ 2 }-px+q=0$ , then find the quadratic equation with the roots $ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) \left( { \alpha }^{ 3 }-{ \beta }^{ 3 } \right) $ and $ { \alpha }^{ 3 }{ \beta }^{ 2 }+{ \alpha }^{ 2 }{ \beta }^{ 3 }$

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

Using Vieta's formulas for x^2 - px + q = 0, we have the sum of roots as alpha + beta = p and the product as alpha*beta = q. We find the first root of the new equation by factoring the expression: (alpha^2 - beta^2)(alpha^3 - beta^3) = (alpha - beta)(alpha + beta) * (alpha - beta)(alpha^2 + alpha*beta + beta^2). Substituting the sum and product gives (p^2 - 4q)(p^2 - q). The second root simplifies as alpha^3*beta^2 + alpha^2*beta^3 = (alpha*beta)^2(alpha + beta) = p*q^2. The new quadratic equation is found using the identity y^2 - (sum of new roots)y + (product of new roots) = 0. The sum of the roots is p*q^2 + (p^2 - 4q)(p^2 - q) and the product is p*q^2 * (p^2 - 4q)(p^2 - q). Therefore, the required equation is y^2 + p{q^2 + (p^2 - 4q)(p^2 - q)}y + p^2*q^2(p^2 - 4q)(p^2 - q) = 0.

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