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Question: If $\alpha, \beta, \gamma$ are the roots of $x^3 + ax^2 + b = 0$, then the value of $\begin{vmatrix}...

If α,β,γ\alpha, \beta, \gamma are the roots of x3+ax2+b=0x^3 + ax^2 + b = 0, then the value of ∣αβγβγαγαβ∣\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} is equal to

A

-a^3

B

a^3 - ab

C

a^3

D

a^2 - 3b

Answer

a^3

Explanation

Solution

The given cubic equation is x3+ax2+b=0x^3 + ax^2 + b = 0. Let the roots of this equation be α,β,γ\alpha, \beta, \gamma. Comparing this equation with the standard cubic equation Ax3+Bx2+Cx+D=0Ax^3 + Bx^2 + Cx + D = 0, we have A=1,B=a,C=0,D=bA=1, B=a, C=0, D=b.

Using Vieta's formulas:

  1. Sum of the roots: α+β+γ=−BA=−a1=−a\alpha + \beta + \gamma = -\frac{B}{A} = -\frac{a}{1} = -a
  2. Sum of the products of the roots taken two at a time: αβ+βγ+γα=CA=01=0\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} = \frac{0}{1} = 0
  3. Product of the roots: αβγ=−DA=−b1=−b\alpha\beta\gamma = -\frac{D}{A} = -\frac{b}{1} = -b

We need to evaluate the determinant:

D=∣αβγβγαγαβ∣D = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}

We can evaluate this determinant using column operations. Add the second and third columns to the first column (C1→C1+C2+C3C_1 \to C_1 + C_2 + C_3):

D=∣α+β+γβγβ+γ+αγαγ+α+βαβ∣D = \begin{vmatrix} \alpha+\beta+\gamma & \beta & \gamma \\ \beta+\gamma+\alpha & \gamma & \alpha \\ \gamma+\alpha+\beta & \alpha & \beta \end{vmatrix}

Now, take out the common factor (α+β+γ)(\alpha+\beta+\gamma) from the first column:

D=(α+β+γ)∣1βγ1γα1αβ∣D = (\alpha+\beta+\gamma) \begin{vmatrix} 1 & \beta & \gamma \\ 1 & \gamma & \alpha \\ 1 & \alpha & \beta \end{vmatrix}

Expand the 3×33 \times 3 determinant:

D=(α+β+γ)[1(γβ−αα)−β(1⋅β−1⋅α)+γ(1⋅α−1⋅γ)]D = (\alpha+\beta+\gamma) [1(\gamma\beta - \alpha\alpha) - \beta(1\cdot\beta - 1\cdot\alpha) + \gamma(1\cdot\alpha - 1\cdot\gamma)]

D=(α+β+γ)[βγ−α2−β2+αβ+αγ−γ2]D = (\alpha+\beta+\gamma) [\beta\gamma - \alpha^2 - \beta^2 + \alpha\beta + \alpha\gamma - \gamma^2]

D=(α+β+γ)[−(α2+β2+γ2−αβ−βγ−γα)]D = (\alpha+\beta+\gamma) [-(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)]

Now, we use the algebraic identity:

α2+β2+γ2−αβ−βγ−γα=(α+β+γ)2−3(αβ+βγ+γα)\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha = (\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)

Substitute this into the expression for DD:

D=(α+β+γ)[−((α+β+γ)2−3(αβ+βγ+γα))]D = (\alpha+\beta+\gamma) [- ((\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha))]

Now, substitute the values from Vieta's formulas:

α+β+γ=−a\alpha+\beta+\gamma = -a

αβ+βγ+γα=0\alpha\beta+\beta\gamma+\gamma\alpha = 0

D=(−a)[−((−a)2−3(0))]D = (-a) [- ((-a)^2 - 3(0))]

D=(−a)[−(a2−0)]D = (-a) [- (a^2 - 0)]

D=(−a)[−a2]D = (-a) [-a^2]

D=a3D = a^3

The value of the determinant is a3a^3.