Question

If the zeros of the polynomial f(x) = $ax^3 + 3bx^2 +3cx + d$ are in AP, prove that $2b^3 - 3abc + a^2d$ = 0

Answer 1

Hint: To solve this problem, one should have theoretical knowledge of cubic equations and the relationship between their coefficients and roots. The formulas that will be used to verify this relationship are-
$$\mathrm\alpha+\mathrm\beta+\mathrm\gamma=\dfrac{-\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{\mathrm\beta\mathrm\gamma}+\mathrm{\mathrm\gamma\mathrm\alpha}=\dfrac{\mathrm c}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta\mathrm\gamma}=\dfrac{-\mathrm d}{\mathrm a}$$

Complete step-by-step answer:
 Let the roots be k - r, k, k + r. By comparing with the f(x), the sum of roots is-
$$\mathrm k-\mathrm r+\mathrm k+\mathrm k+\mathrm r=-\dfrac{3\mathrm b}{\mathrm a}\\3\mathrm k=-\dfrac{3\mathrm b}{\mathrm a}\\\mathrm k=-\dfrac{\mathrm b}{\mathrm a}$$
By comparing with the f(x), the sum of the product of roots is-
$$\mathrm k\left(\mathrm k-\mathrm r\right)+\mathrm k\left(\mathrm k+\mathrm r\right)+\left(\mathrm k-\mathrm r\right)\left(\mathrm k+\mathrm r\right)=\dfrac{\mathrm c}{\mathrm a}\\\mathrm k^2-\mathrm{kr}+\mathrm k^2+\mathrm{kr}+\mathrm k^2-\mathrm r^2=\dfrac{\mathrm c}{\mathrm a}\\3\mathrm k^2-\mathrm r^2=\dfrac{\mathrm c}{\mathrm a}\\\dfrac{3\mathrm b^2}{\mathrm a^2}-\mathrm r^2=\dfrac{\mathrm c}{\mathrm a}\\\mathrm r^2=\dfrac{3\mathrm b^2-\mathrm{ac}}{\mathrm a^2}$$
By comparing with the f(x), the product of roots is-
$$\mathrm k\left(\mathrm k-\mathrm r\right)\left(\mathrm k+\mathrm r\right)=-\dfrac{\mathrm d}{\mathrm a}\\\mathrm k\left(\mathrm k^2-\mathrm r^2\right)=-\dfrac{\mathrm d}{\mathrm a}\\-\dfrac{\mathrm b}{\mathrm a}\left(\dfrac{\mathrm b^2}{\mathrm a^2}+\dfrac{\mathrm{ac}-3\mathrm b^2}{\mathrm a^2}\right)=-\dfrac{\mathrm d}{\mathrm a}\\\dfrac{-2\mathrm b^3+3\mathrm{abc}}{\mathrm a^2}=\mathrm d\\2\mathrm b^3-3\mathrm{abc}+\mathrm a^2\mathrm d=0$$
Hence, proved.
Note: In such types of questions be careful in what we assume the roots. In this question, we assumed the roots as k-r, k and k+r so that d gets cancelled and we get the value of a in the first equation itself. This makes the calculation a lot easier.