uation x^3-3px^2+3qx-r=0 has equal pair of roots,show that ((3p^3-4pq+r)/(p^2-q))

Question

equation x^3-3px^2+3qx-r=0 has equal pair of roots,show that ((3p^3-4pq+r)/(p^2-q))

Answer 1

There is no equation to prove, just an expression involving p, q and r. Therefore a full solution is not possible.

Let the roots be a and b then:

(x-a)²(x-b)=(x²-2ax+a²)(x-b)=x³-(2a+b)x²+(a²+2ab)x-a²b.

Equating coefficients:

constant: r=a²b, b=r/a²

x term: 3q=a²+2ab=a²+2r/a, 3aq=a³+2r, a³=3aq-2r, r=(3aq-a³)/2;

x² term: 3p=2a+b=2a+r/a², 3a²p=2a³+r=6aq-3r, r=3a²p-2a³;

(3q-a²)/2=3ap-2a²

3a²p-6aq+3r=0, a²p-2aq+r=0, a=(2q±√(4q²-4pr))/2p, a=(q±√(q²-pr))/p.

b=r/a²=p²r/(2q²±2q√(q²-pr)-pr).