Solveeit Logo
Login

Question

Question: If $\alpha, \beta, \gamma$ are the roots of the equation $4x^3 - 3x^2 + 2x - 1 = 0$, then $\alpha^3 ...

If α,β,γ\alpha, \beta, \gamma are the roots of the equation 4x3−3x2+2x−1=04x^3 - 3x^2 + 2x - 1 = 0, then α3+β3+γ3=\alpha^3 + \beta^3 + \gamma^3 =

Answer

3/64

Explanation

Solution

Given the cubic equation:

4x3−3x2+2x−1=04x^3 - 3x^2 + 2x - 1 = 0

Divide by 4 to obtain a monic polynomial:

x3−34x2+12x−14=0x^3 - \frac{3}{4}x^2 + \frac{1}{2}x - \frac{1}{4} = 0

Let the roots be α\alpha, β\beta, γ\gamma. By Vieta's formulas, we have:

α+β+γ=34,αβ+βγ+γα=12,αβγ=14\alpha + \beta + \gamma = \frac{3}{4}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{2}, \quad \alpha\beta\gamma = \frac{1}{4}

Use the identity for the sum of cubes:

α3+β3+γ3=(α+β+γ)3−3(α+β+γ)(αβ+βγ+γα)+3αβγ\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)^3 - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma

Substitute the values:

α3+β3+γ3=(34)3−3(34)(12)+3(14)\alpha^3 + \beta^3 + \gamma^3 = \left(\frac{3}{4}\right)^3 - 3\left(\frac{3}{4}\right)\left(\frac{1}{2}\right) + 3\left(\frac{1}{4}\right)

Calculate each term:

(34)3=2764,3(34)(12)=98=7264,3(14)=34=4864\left(\frac{3}{4}\right)^3 = \frac{27}{64}, \quad 3\left(\frac{3}{4}\right)\left(\frac{1}{2}\right) = \frac{9}{8} = \frac{72}{64}, \quad 3\left(\frac{1}{4}\right) = \frac{3}{4} = \frac{48}{64}

Thus,

α3+β3+γ3=2764−7264+4864=27−72+4864=364\alpha^3 + \beta^3 + \gamma^3 = \frac{27}{64} - \frac{72}{64} + \frac{48}{64} = \frac{27 - 72 + 48}{64} = \frac{3}{64}