Question

Let α, β, γ be the three roots of the equation x³+bx+c=0. If βγ =1=-α, then b³+2c³-3α³-6β³-8γ³ is equal to

• A. \(\frac{169}{8}\)
• B. \(\frac{155}{8}\)
• C. 19
• D. 28

Hint:

Cube roots of unity satisfy \(\omega^3 = 1\) and simplify symmetric expressions for roots of equations.

Answer 1

The Correct Option is C

Step 1: Use the properties of roots.
- Given that the roots are \(\alpha, \beta, \gamma\), we know from Vieta’s formulas: \[ \alpha + \beta + \gamma = 0, \quad \alpha\beta + \beta\gamma + \gamma\alpha = b, \quad \alpha\beta\gamma = -c. \] - Substituting \(\alpha = -1\) and \(\beta\gamma = 1\): \[ -1 + \beta + \gamma = 0 \implies \beta + \gamma = 1. \] - Substituting into \(\alpha\beta\gamma = -c\): \[ (-1)(\beta\gamma) = -c \implies -1 = -c \implies c = 1. \] Step 2: Solve for \(b\).
- Substituting \(\alpha = -1\), \(\beta\gamma = 1\), and \(c = 1\) into \(\alpha\beta + \beta\gamma + \gamma\alpha = b\): \[ (-1)\beta + 1 + \gamma(-1) = b \implies -\beta - \gamma + 1 = b \implies -(1) + 1 = b \implies b = 0. \] Step 3: Evaluate the given expression.
- Substituting \(b = 0\) and \(c = 1\): \[ b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3 = 0^3 + 2(1)^3 - 3(-1)^3 - 6\beta^3 - 8\gamma^3. \] \[ = 0 + 2 + 3 - 6\beta^3 - 8\gamma^3. \] - Using \(\beta\gamma = 1\), we know \(\beta^3\gamma^3 = (\beta\gamma)^3 = 1^3 = 1\). Let \(\beta^3 = x\) and \(\gamma^3 = \frac{1}{x}\). Substituting: \[ 2 + 3 - 6x - \frac{8}{x}. \] Step 4: Solve for \(\beta\) and \(\gamma\).
- Using \(\beta + \gamma = 1\) and \(\beta\gamma = 1\), the roots of the quadratic equation \(t^2 - t + 1 = 0\) are \(\beta = -\omega\) and \(\gamma = -\omega^2\), where \(\omega\) is a cube root of unity. Substituting these values into the expression simplifies the result: \[ \boxed{19}. \] Final Answer: The value is \(19\).