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Question: Sigma (ab)^ 3 for a cubic equation x^3+4x+1...

Sigma (ab)^ 3 for a cubic equation x^3+4x+1

Answer

67

Explanation

Solution

The problem asks for the value of ∑(αβ)3\sum (\alpha\beta)^3 for the cubic equation x3+4x+1=0x^3+4x+1=0. Let the roots of the equation be α,β,γ\alpha, \beta, \gamma.

1. Apply Vieta's Formulas: For a cubic equation ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0, Vieta's formulas state:

  • Sum of roots: α+β+γ=−b/a\alpha+\beta+\gamma = -b/a
  • Sum of products of roots taken two at a time: αβ+βγ+γα=c/a\alpha\beta+\beta\gamma+\gamma\alpha = c/a
  • Product of roots: αβγ=−d/a\alpha\beta\gamma = -d/a

For the given equation x3+0x2+4x+1=0x^3+0x^2+4x+1=0, we have a=1,b=0,c=4,d=1a=1, b=0, c=4, d=1. So,

  • α+β+γ=−0/1=0\alpha+\beta+\gamma = -0/1 = 0
  • αβ+βγ+γα=4/1=4\alpha\beta+\beta\gamma+\gamma\alpha = 4/1 = 4
  • αβγ=−1/1=−1\alpha\beta\gamma = -1/1 = -1

2. Define New Variables: We need to find the value of ∑(αβ)3=(αβ)3+(βγ)3+(γα)3\sum (\alpha\beta)^3 = (\alpha\beta)^3 + (\beta\gamma)^3 + (\gamma\alpha)^3. Let A=αβA = \alpha\beta, B=βγB = \beta\gamma, C=γαC = \gamma\alpha. We need to find A3+B3+C3A^3+B^3+C^3.

3. Calculate Sums and Products of New Variables:

  • Sum:
    A+B+C=αβ+βγ+γα=4A+B+C = \alpha\beta+\beta\gamma+\gamma\alpha = 4

  • Sum of products taken two at a time:
    AB+BC+CA=(αβ)(βγ)+(βγ)(γα)+(γα)(αβ)AB+BC+CA = (\alpha\beta)(\beta\gamma) + (\beta\gamma)(\gamma\alpha) + (\gamma\alpha)(\alpha\beta)
    =αβ2γ+βγ2α+γα2β= \alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta
    Factor out αβγ\alpha\beta\gamma:
    =αβγ(β+γ+α)= \alpha\beta\gamma(\beta+\gamma+\alpha)
    Substitute the values from Vieta's formulas:
    =(−1)(0)=0= (-1)(0) = 0

  • Product:
    ABC=(αβ)(βγ)(γα)ABC = (\alpha\beta)(\beta\gamma)(\gamma\alpha)
    =α2β2γ2=(αβγ)2= \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2
    Substitute the value from Vieta's formulas:
    =(−1)2=1= (-1)^2 = 1

4. Use the Sum of Cubes Identity: The identity for the sum of cubes of three variables is:
A3+B3+C3−3ABC=(A+B+C)(A2+B2+C2−(AB+BC+CA))A^3+B^3+C^3 - 3ABC = (A+B+C)(A^2+B^2+C^2 - (AB+BC+CA))

First, calculate A2+B2+C2A^2+B^2+C^2:
A2+B2+C2=(A+B+C)2−2(AB+BC+CA)A^2+B^2+C^2 = (A+B+C)^2 - 2(AB+BC+CA)
Substitute the values calculated above:
A2+B2+C2=(4)2−2(0)A^2+B^2+C^2 = (4)^2 - 2(0)
=16−0=16= 16 - 0 = 16

Now, substitute all values into the sum of cubes identity:
A3+B3+C3−3(1)=(4)(16−0)A^3+B^3+C^3 - 3(1) = (4)(16 - 0)
A3+B3+C3−3=(4)(16)A^3+B^3+C^3 - 3 = (4)(16)
A3+B3+C3−3=64A^3+B^3+C^3 - 3 = 64
A3+B3+C3=64+3A^3+B^3+C^3 = 64 + 3
A3+B3+C3=67A^3+B^3+C^3 = 67

Thus, ∑(αβ)3=67\sum (\alpha\beta)^3 = 67.