lf the roots  of the equation  x3−px2+qx−r=0 are in H.P., then

The equation whose roots are reciprocals of the roots of the given equation is given by 1x3−px2+qx−r=0⇒rx3−qx2+px−1=0Since the roots of the given equation are in H.P. Therefore, the roots of equation (i) are in A.P. Let its roots be a - d, a and a+ d. ThenThen,(a-d)+a+(a+d)=−−1r⇒3a=qr⇒a=93rSince a is a root of equation { i).∴ ra3−qa2+pa−1=0⇒ rq3r3−qq3r2+pq3r−1=0⇒ q327r2−q39r2+pq3r−1=0⇒ q3−3q3+9pqr−27r2=0⇒27r2−9pqr+2q3=0