Given equation, x3+qx+r=0………………(1)
bc+cb=b2+c2bc=aa2+b2+c2−a2abc=a(∑a)2−2∑ab−a2abc=a(a2+2qr)
⟹y=x(x2+2qr)
⟹x3+2qx–ry=0……………..(2)
Subtracting (1) from (2), we have
qx–ry−r=0⟹x=(rq)(1+y)……………..(3)
Substituting value of x from (3) in (1), we get
(rq)3(1+y)3+r(qq)(1+y)+q3=0
⟹r2y3+3r2y2+(2r2+q3)y+(r2+2q3)=0