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).