Question

If the zeroes of polynomial x³-ax²+bx-c are in AP then show that 2a³-9ab+27c=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:zeroes \: of \: {x}^{3} - {ax}^{2} + bx - c \: are \: in \: AP$

Let assume that,

$\rm :\longmapsto\:f(x) \: = \: {x}^{3} - {ax}^{2} + bx - c \:$

and

Let further assume that,

$\rm :\longmapsto\:zeroes \: of \: f(x) \: be \: y - z, \: y, \: y + z$

as they are in AP.

Now,

we further know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:y - z + y + y + z = - \: \dfrac{( - a)}{1}$

$\rm :\longmapsto\:3y = a$

$\bf\implies \:y = \dfrac{a}{3}$

Now, since y is zero of f(x),

We have,

$\rm :\longmapsto\:f(x) \: = \: {x}^{3} - {ax}^{2} + bx - c \:$

So,

$\rm :\longmapsto\:f(y) = 0$

$\rm :\longmapsto\:f\bigg(\dfrac{a}{3} \bigg) = 0$

$\rm :\longmapsto\:\bigg(\dfrac{a}{3} \bigg)^{3} - a\bigg(\dfrac{a}{3} \bigg) ^{2} + b\bigg(\dfrac{a}{3} \bigg) - c = 0$

$\rm :\longmapsto\:\dfrac{ {a}^{3} }{27} - \dfrac{ {a}^{3} }{9} + \dfrac{ab}{3} - c = 0$

$\rm :\longmapsto\:\dfrac{ {a}^{3} - 3 {a}^{3} + 9ab - 27c}{27} = 0$

$\rm :\longmapsto\:\dfrac{ - 2 {a}^{3} + 9ab - 27c}{27} = 0$

$\rm :\longmapsto\: - 2 {a}^{3} + 9ab - 27c = 0$

$\rm :\longmapsto\: 2 {a}^{3} - 9ab + 27c = 0$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\: \alpha,\beta, \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then$

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \: \dfrac{b}{a} }}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}$

$\boxed{ \sf{ \: \alpha \beta \gamma = - \: \dfrac{d}{a} }}$