If $α, β, γ$ are the zeroes of the cubic polynomial $x^{3} + 4 x + 2$ , then find the value of:
$\frac{1}{α + β} + \frac{1}{β + γ} + \frac{1}{γ + α}$

Open in App

Solution

We have,
$P (x) = x^{3} + 4 x + 2$

Since, $α, β, γ$ are the zeroes of this polynomial.

Now,
$α + β + γ = - \frac{0}{1} = 0$

$α β + β γ + γ α = \frac{4}{1} = 4$

$α β γ = - \frac{2}{1} = - 2$

Since,
$\frac{1}{α + β} + \frac{1}{β + γ} + \frac{1}{γ + α}$

$\Rightarrow \frac{(β + γ) (γ + α) + (α + β) (γ + α) + (α + β) (β + γ)}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow \frac{β γ + α β + γ^{2} + α γ + α γ + α^{2} + β γ + α β + α β + α γ + β^{2} + β γ}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow \frac{α^{2} + β^{2} + γ^{2} + 3 (α β + β γ + γ α)}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow \frac{α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α) + (α β + β γ + γ α)}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow \frac{(α + β + γ)^{2} + (α β + β γ + γ α)}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow \frac{0 + 0}{(α + β) (β + γ) (γ + α)}$

$\Rightarrow 0$

Hence, this is the answer.

Suggest Corrections

Similar questions

α, β, γ

P (x) = x^{3} + 4 x + 2

\frac{1}{α + β} + \frac{1}{β + γ} + \frac{1}{γ + α}

α, β, γ

x^{3} + 4 x + 1 = 0

(α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1}

If α, β, γ are the roots of the equation $x^{3} + 4 x + 1 = 0$ ,then $(α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1} =$

α

β

γ

x^{3} + p x^{2} + q x^{2} + 2

α β + 1 = 0

2 p + q + 5

α, β, γ

x^{3} - x^{2} - 1 = 0

\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}

Join BYJU'S Learning Program