Answer to Question #116896 in Algebra for jay

Question

4. If the roots of the cubic az
+ bz + cz + d = 0 form an arithmetic progression α − β,
2
2
α, α + β, prove that (2b
− 9ac)b + 27a
d = 0.

Answer 1

We are given that $az^3+bz^2+cz+d=0$  has three roots in arithmetic progression

$\alpha-\beta,\alpha,\alpha+\beta.$

We know that

$$z_1+z_2+z_3=-{b \over a}$$ $$z_1z_2z_3=-{d \over a}$$ $$z_1z_2+z_2z_3+z_1z_3={c \over a}$$

Then

$$\alpha-\beta+\alpha+\alpha+\beta=-{b \over a}$$

$$(\alpha-\beta)(\alpha)(\alpha+\beta)=-{d \over a}$$

$$(\alpha-\beta)(\alpha)+(\alpha)(\alpha+\beta)+(\alpha-\beta)(\alpha+\beta)={c \over a}$$

We have

$$\alpha=-{b \over 3a}$$

$$\alpha(\alpha^2-\beta^2)=-{d \over a}$$

$$3\alpha^2-\beta^2={c \over a}$$

$$\beta^2=3\alpha^2-{c \over a}={b^2 \over 3a^2}-{c \over a}$$

Then

$$-{b \over 3a}({b^2 \over 9a^2}-{b^2 \over 3a^2}+{c \over a})=-{d \over a}$$

$${b(2b^2-9ac )\over 9a^2}=-3d$$

$$(2b^2-9ac )b+27a^2d=0$$