Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \( 4x^3 - 3x^2 + 2x - 1 = 0 \), then \( \alpha^3 + \beta^3 + \gamma^3 = \)

• A. \(\frac{2}{27}\)
• B. \(\frac{1}{8}\)
• C. \(\frac{3}{64}\)
• D. \(\frac{27}{128}\)

Hint:

Always double-check algebraic manipulations and identity applications when dealing with sums of powers of polynomial roots.

Answer 1

The Correct Option is C

Step 1: Use the identity for the sum of cubes. The identity for the sum of cubes of the roots of a cubic polynomial \(ax^3 + bx^2 + cx + d = 0\) is: \[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha \beta \gamma - (\alpha + \beta + \gamma)(\alpha \beta + \beta \gamma + \gamma \alpha) + (\alpha + \beta + \gamma)^3 \] Step 2: Apply Vieta's formulas. From Vieta's formulas, we know: \[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-3}{4} = \frac{3}{4}, \quad \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = \frac{2}{4} = \frac{1}{2}, \quad \alpha \beta \gamma = -\frac{d}{a} = -\frac{-1}{4} = \frac{1}{4} \] Step 3: Substitute values into the identity. Substitute the known values into the identity: \[ \alpha^3 + \beta^3 + \gamma^3 = 3\left(\frac{1}{4}\right) - \left(\frac{3}{4}\right)\left(\frac{1}{2}\right) + \left(\frac{3}{4}\right)^3 \] Calculate each term: \[ = \frac{3}{4} - \frac{3}{8} + \frac{27}{64} \] \[ = \frac{24}{64} - \frac{24}{64} + \frac{27}{64} = \frac{27}{64} \] However, since the correct answer is given as \(\frac{3}{64}\), check for potential errors or misinterpretations: Correct the initial identity use, as it seems the derived formula was incorrect. The actual identity without expanding \( (\alpha + \beta + \gamma)^3 \) is: \[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha \beta \gamma + (\alpha + \beta + \gamma)^3 - 3(\alpha + \beta + \gamma)(\alpha \beta + \beta \gamma + \gamma \alpha) \] Substituting back, with correct computation: \[ \alpha^3 + \beta^3 + \gamma^3 = 3\left(\frac{1}{4}\right) + \left(\frac{3}{4}\right)^3 - 3\left(\frac{3}{4}\right)\left(\frac{1}{2}\right) \] \[ = \frac{3}{4} + \frac{27}{64} - \frac{9}{8} \] \[ = \frac{48}{64} + \frac{27}{64} - \frac{72}{64} = \frac{3}{64} \]