Let p and q be real numbers such that p≠0,p3 ≠q and p3≠ -q. If α and ß are nonzero complex numbers satisfying α + ß = - p and α3+ß3=q, then a quadratic equation having α/ß and ß/α as its roots is
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(p3+q) x2-(p3+2q)x+(p3+q)=0
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(p3+q) x2- (p3-2q)x+(p3+q)=0
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(p3-q) x2- (5p3-2q)x+(p3-1)=0
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(p3-q) x2- (5p3+2q)x+(p3-q)=0
Given alpha+beta = -p and alpha^3+beta^3 = q. We know alpha^3+beta^3 = (alpha+beta)((alpha+beta)^2 - 3*alpha*beta) = q. So -p(p^2 - 3*alpha*beta) = q. Thus, -p^3 + 3*p*alpha*beta = q, so alpha*beta = (p^3+q)/(3p). The roots are alpha/beta and beta/alpha. Their sum = (alpha^2+beta^2)/(alpha*beta) = ((alpha+beta)^2 - 2*alpha*beta)/(alpha*beta) = (p^2 - 2*alpha*beta)/(alpha*beta) = p^2/(alpha*beta) - 2. Their product = 1. Substituting alpha*beta, the quadratic is x^2 - (sum)x + 1 = 0.
Using the identity for the sum of cubes, we have alpha cubed plus beta cubed equals (alpha plus beta) multiplied by (alpha squared minus alpha beta plus beta squared). Substituting the given values, q equals (-p) multiplied by (p squared minus 3 alpha beta), which simplifies to alpha beta equals (p cubed plus q) divided by (3p). The sum of the new roots is (alpha divided by beta) plus (beta divided by alpha), which equals (alpha squared plus beta squared) divided by (alpha beta). This simplifies to ((alpha plus beta) squared minus 2 alpha beta) divided by (alpha beta), yielding (p cubed minus 2q) divided by (p cubed plus q). The product of the new roots is 1, which equals (p cubed plus q) divided by (p cubed plus q). Using these sums and products to form the quadratic equation gives (p cubed plus q) x squared minus (p cubed minus 2q) x plus (p cubed plus q) equals 0.