Given equation, x3+qx+r=0………………(1)
b+ca2=a+b+c−aa2=−1a
⟹y=−1x⟹x=−1y………….(2)
Substituting value of x from (2) in (1), we get
(−1y)3+q(−1y)+r=0
⟹ry3−qy2−1=0