ax^3 + 3bx^2 + 3cx + d = 0 are in G.P
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Let roots be k/p, k and kp which are in GP
Sum of roots =k/p+k+kp =-3b/a (A)
Sum of products in pairs=k^2/p+k^2p+k^2 = 3c/a (B)
Product =k^3=-d/a (C)
Now eliminate k and p from the relations A,B and C
and you get ac^3=bd^3
Sum of roots =k/p+k+kp =-3b/a (A)
Sum of products in pairs=k^2/p+k^2p+k^2 = 3c/a (B)
Product =k^3=-d/a (C)
Now eliminate k and p from the relations A,B and C
and you get ac^3=bd^3