Q54 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

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

  • A
    $\frac{2}{27}$
  • B
    $\frac{1}{8}$
  • Correct answer
    C
    $\frac{3}{64}$
  • D
    $\frac{27}{128}$

Solution

$\alpha, \beta, \gamma$ are the roots of the equation $\begin{aligned} & 4 x^3-3 x^2+2 x-1=0 \\ & \Rightarrow 4 \alpha^3=3 \alpha^2-2 \alpha+1 \Rightarrow 4 \beta^3=3 \beta^2-2 \beta+1 \\ & \Rightarrow 4 \gamma^3=3 \gamma^2-2 \gamma+1 \\ & 4\left(\alpha^3+\beta^3+\gamma^3\right)=3\left(\alpha^2+\beta^2+\gamma^2\right)-2(\alpha+\beta+\gamma)+3 \\ & =3\left[(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\alpha \gamma)\right]-2 \times \frac{3}{4}+3 \\ & =3\left(\frac{9}{16}-2 \times \frac{1}{2}\right)-\frac{3}{2}+3=\frac{3}{2}-\frac{21}{16}=\frac{3}{16} \\ & \Rightarrow \alpha^3+\beta^3+\gamma^3=\frac{3}{64} . \end{aligned}$

Hence, the correct answer is option: 'C'.

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