If the roots of the cubic equation ax^3+bx^2+cx+d=0 are in G.P then prove c^3a=b^3d

let roots be m,n,p m+n+p=-b/a,mnp=-d/a,mn+np+mp=c/a,roots are in gp implies n^2=mp ,mpn=-d/a impliesn^2*n=-d/a implies n=(-d/a)^1/3 and mn+np+mp=c/a implies mn+np+n^2=c/a implies n(m+n+p)=-b/a implies n(-b/a)=c/a implies n=-c/b previously n=(-d/a)^1/3 cube this and equate values of n in both the cases

Since the roots are in g.p., let them be k, kr, kr^2. We know that the sum of the roots of a cubic equations equals -b/a. This means k + kr + kr^2 = -b/a. Let that be equations number 1. We also know that k(kr) + k(kr^2) = kr(kr^2) = c/a , which means (k^2)r + (k^2)r^2 + (k^2)r^3 = c/a, which implies kr( k + kr + kr^2) = c/a [ Taking kr common ] . Now, substituting equation 1 into this equation, we get  kr(-b/a) = c/a, which implies that kr = -c/b [ the a’s get cancelled out ] . Let this be equation number 2. Now, the products of the roots of a cubic equation = -d/a , this implis that k(kr)(kr^2) = -d/a, which implies (kr)^3 = -d/a . Substituting equation 2 in this equation, we get (-c/b)^3 = -d/a, which implies that      -c^3/b^3 =-d/a, and (-c^3)a = (-b^3)d, or (c^3)a = (b^3)d. Hence proved