# If the roots of the cubic equation ax^3+bx^2+cx+d=0 are in G.P then which of the following is correct ? A) (c^3)a=(b^3)d; B) c(a^3)=b(d^3); C) (a^3)b=(c^3)d; D) a(b^3)=c(d^3)

## b

Mar 21, 2018

${c}^{3} a = {b}^{3} d$

#### Explanation:

If $\alpha$, $\beta$ and $\gamma$ are the three roots then we must have

$a {x}^{3} + b {x}^{2} + c x + d \equiv a \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$

comparing coefficients of various powers of $x$ on both sides leads to

$\alpha + \beta + \gamma = - \frac{b}{a} q \quad q \quad \quad \left[1\right]$
$\alpha \beta + \beta \gamma + \gamma \alpha = + \frac{c}{a} q \quad \left[2\right]$
$\alpha \beta \gamma = - \frac{d}{a} q \quad q \quad q \quad q \quad q \quad \left[3\right]$

In this problem, the three roots are in GP, so that $\beta = \alpha r$ and gamma = alpha r^2#. Substituting this in [1]. [2] and [3] gives

$\alpha \left(1 + r + {r}^{2}\right) = - \frac{b}{a} q \quad q \quad \quad \left[1 a\right]$
${\alpha}^{2} \left(r + {r}^{2} + {r}^{3}\right) = + \frac{c}{a} q \quad q \quad \left[2 a\right]$
${\alpha}^{3} {r}^{3} = - \frac{d}{a} q \quad q \quad q \quad q \quad q \quad q \quad \left[3 a\right]$

Divideing [2a] by [1a] leads to $\alpha r = - \frac{c}{b}$ and substituting this in [3a] gives

${\left(- \frac{c}{b}\right)}^{3} = - \frac{d}{a} \implies - {c}^{3} / {b}^{3} = - \frac{d}{a} \implies {c}^{3} a = {b}^{3} d$