Q73 / 200 · Manipal (MET) 2010 · Mathematics

If $\alpha, \beta, \gamma$ are such that $\alpha+\beta+\gamma=2$, $\alpha^2+\beta^2+\gamma^2=6, \quad \alpha^3+\beta^3+\gamma^3=8, \quad$ then $\alpha^4+\beta^4+\gamma^4$ is

  • A
    5
  • Correct answer
    B
    18
  • C
    12
  • D
    36

Solution

We have, $(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)$ $\Rightarrow 4=6+2(\alpha \beta+\beta \gamma+\gamma \alpha)$ $\Rightarrow \beta \gamma+\gamma \alpha+\alpha \beta=-1...(i)$ Also, $\alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma$ $=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)$ $\Rightarrow 8-3 \alpha \beta \gamma=2(6+1)$ $\Rightarrow 3 \alpha \beta \gamma=8-14=-6$ or, $\alpha \beta \gamma=-2...(ii)$ Now, $\begin{aligned} \left(\alpha^2+\beta^2+\gamma^2\right)^2 & =\alpha^4+\beta^4+\gamma^4 +2\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\right) \end{aligned}$ $\begin{aligned} &=\left(\alpha^4+\beta^4+\gamma^4\right)+2\left[(\alpha \beta+\beta \gamma+\gamma \alpha)^2\right. -2 \alpha \beta \gamma(\alpha+\beta+\gamma)]\end{aligned}$ $\begin{aligned} \Rightarrow\left(\alpha^4+\beta^4+\gamma^4\right)= & 36 -2\left[(-1)^2-2(-2)(2)\right]=18\end{aligned}$

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

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