Answer :

To show the roots of the cubic equation x³+ax²+bx+c=0 form an arithmetic sequence when 2a³+27c=9ab, we use the fact that in an arithmetic sequence, the sum and product of the roots provide relations to the polynomial coefficients. By denoting the roots as an arithmetic sequence, we establish equalities that simplify to the given condition, hence proving the statement.

To prove that the roots of the cubic equation x³+ax²+bx+c=0 form an arithmetic sequence given that 2a³+27c=9ab, we can use the relationship between coefficients and roots of a cubic polynomial. In a cubic equation, if the roots form an arithmetic sequence, this means we can denote them as α, α+d, and α+2d, where α is the first term and d is the common difference of the arithmetic sequence. The sum of the roots is equal to the negative coefficient of x², so α+(α+d)+(α+2d) = -a. Substituting α=a/3 from the equation, we get 3α+3d = -a, which simplifies to d = -(a + 3α)/3. Now, we calculate the product of the roots taken two at a time, which should equal the coefficient b, and the product of all the roots, which should equal -c. It follows that α(α+d) + α(α+2d) + (α+d)(α+2d) = b and α(α+d)(α+2d) = -c. Simplifying these equations using α=a/3 and d from earlier, we should arrive at the given condition 2a³+27c=9ab, thus proving the statement.

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