Given equation, x3+qx+r=0………………(1)
b2c2=a2b2c2a2=r2a2
⟹y=r2x2⟹x=r√y………….(2)
Substituting value of x from (2) in (1), we get
(r√y)3+q(r√y)+r=0
⟹r√y3+qry+r3=0⟹√y3=−(qy+r2)=0
y3=q2y2+2qr2y+r4[Taking squares on both sides]
⟹y3−q2y2−2qr2y−r4=0