MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Global Maxima

If α, β, γ ...

Question

If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ , then the value of $(1 - α) (1 - β) (1 - γ)$ is:

A

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

- 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 1$
The equation is $: x^{3} - 3 x^{2} + 2 x - 1 = 0$

By comparing and using the relation between zeroes and roots

$(1 - α) (1 - β) (1 - γ)$

$= 1 - (α + β + γ) + (α β + β γ + γ α) - α β γ = 1 - (\frac{3}{1}) + (\frac{2}{1}) - (\frac{1}{1}) = - 1$

Suggest Corrections

thumbs-up

1

Similar questions

α

β

γ

are the roots of

x^{3} - x^{2} + 8 x + 6 = 0

(1 + α) (1 + β) (1 + γ) =

α, β, γ

are the roots of

x^{3} + p x^{2} + q = 0

q \neq 0

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 / α & 1 / β & 1 / γ 1 / β & 1 / γ & 1 / α 1 / γ & 1 / α & 1 / β \end{matrix} ∣ ∣ ∣ ∣ ∣

α

β

γ

are the roots of

x^{3} + 3 x^{2} - 4 x + 2 = 0

then the equation whose roots are

\frac{1}{α β}

\frac{1}{β γ}

\frac{1}{γ α}

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

α, β, γ

x^{3} - 3 x^{2} + 3 x + 26 = 0

ω

is cube roots of unity then the value of

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Extrema

MATHEMATICS

Watch in App

explore_more_icon

Explore more

COMEDK Engineering

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon