If #alpha#, #beta# and #gamma# are the three roots then we must have

#ax^3+bx^2+cx+d equiv a(x-alpha)(x-beta)(x-gamma)#

comparing coefficients of various powers of #x# on both sides leads to

#alpha+beta+gamma = -b/a qquad qquad quad [1]#

#alpha beta +beta gamma + gamma alpha = +c/a qquad [2]#

#alpha beta gamma = -d/a qquad qquad qquad qquad qquad [3]#

In this problem, the three roots are in GP, so that #beta = alpha r# and gamma = alpha r^2#. Substituting this in [1]. [2] and [3] gives

#alpha(1+r+r^2)= -b/a qquad qquad quad [1a]#

#alpha^2(r+r^2+r^3)= +c/a qquad qquad [2a]#

#alpha^3r^3 = -d/a qquad qquad qquad qquad qquad qquad [3a]#

Divideing [2a] by [1a] leads to #alpha r = -c/b# and substituting this in [3a] gives

#(-c/b)^3=-d/a implies -c^3/b^3=-d/a implies c^3a=b^3d #