Given (abc) 2/3 = . ⇒ (ab + bc + ac) 3 = abc (a + b + c) 3 .....(i)

Now consider the polynomial p(x) = x 3 + mx 2 + nx + p, with roots a, b, c, then have

a + b + c = –m ; ab + bc + ca = n ; abc = – p

using these values equation (i) becomes n 3 = m 3 p ......(ii)

Hence, if m ≠ 0, then equation p(x) = 0 can be written as x 3 + mx 2 + nx + = 0

or m 3 x 3 + m 4 x 2 + nm 3 x + n 3 = 0 ⇒ (mx + n) (m 2 x 2 + (m 3 – mn) x + n 2 ) = 0

It follows that one of the roots of p(x) = 0 is x 1 = – and other two satisfy the condition x 2 x 3 =

⇒ x 12 = x 2 x 3

Thus the roots are the terms of a geometric sequence.

If should be noted that m, n ≠ 0 as in this case x 3 + p = 0 cannot have three real roots.

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