Question

If $\alpha ,\beta ,\gamma$ are such that $\alpha +\beta +\gamma =2,{\alpha }^2+{\beta }^2+{\gamma }^2=6,{\alpha }^3+{\beta }^3+{\gamma }^3=8$, then ${\alpha }^4+{\beta }^4+{\gamma }^4$is equal to _______.

• A. $10$
• B. $12$
• C. $18$
• D. None of these

Answer 1

The correct option is C $18$

We have given that,

$\alpha +\beta +\gamma =2$

${\alpha }^2+{\beta }^2+{\gamma }^2=6$

${\alpha }^3+{\beta }^3+{\gamma }^3=8$

Now, ${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\alpha \gamma )$

$\therefore {\left(2\right)}^2=6+2(\alpha \beta +\beta \gamma +\alpha \gamma )$

$\therefore 4=6+2(\alpha \beta +\beta \gamma +\alpha \gamma )$

$\therefore \alpha \beta +\beta \gamma +\alpha \gamma =\frac{4-6}{2}=\frac{-2}{2}=-1$ Equation (1)

Now, ${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\alpha \gamma )$

$\therefore {\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-(\alpha \beta +\beta \gamma +\alpha \gamma \left)\right)$

$\therefore 8-3\alpha \beta \gamma =2(6-(-1\left)\right)$ [From equation (1)]

$\therefore 8-3\alpha \beta \gamma =14$

$\therefore 3\alpha \beta \gamma =-6$

$\therefore \alpha \beta \gamma =-2$ Equation (2)

Now, ${({\alpha }^2+{\beta }^2+{\gamma }^2)}^2={\left({\alpha }^2\right)}^2+{\left({\beta }^2\right)}^2+{\left({\gamma }^2\right)}^2+2\left({\alpha }^2\right)\left({\beta }^2\right)+2\left({\beta }^2\right)\left({\gamma }^2\right)+2\left({\alpha }^2\right)\left({\gamma }^2\right)$

${({\alpha }^2+{\beta }^2+{\gamma }^2)}^2={\alpha }^4+{\beta }^4+{\gamma }^4+2({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\alpha }^2{\gamma }^2)$

$\therefore {({\alpha }^2+{\beta }^2+{\gamma }^2)}^2={\alpha }^4+{\beta }^4+{\gamma }^4+2[{(\alpha \beta +\beta \gamma +\alpha \gamma )}^2-2\alpha {\beta }^2\gamma -2\alpha \beta {\gamma }^2-2{\alpha }^2\beta \gamma ]$

$\therefore {({\alpha }^2+{\beta }^2+{\gamma }^2)}^2={\alpha }^4+{\beta }^4+{\gamma }^4+2[{(\alpha \beta +\beta \gamma +\alpha \gamma )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )]$

$\therefore {\left(6\right)}^2={\alpha }^4+{\beta }^4+{\gamma }^4+2[{(-1)}^2-2(-2\left)\left(2\right)\right]$ [From Equation (2)]

$\therefore 36={\alpha }^4+{\beta }^4+{\gamma }^4+2[1+8]$

$\therefore 36={\alpha }^4+{\beta }^4+{\gamma }^4+18$

$\therefore {\alpha }^4+{\beta }^4+{\gamma }^4=18$

Thus, answer is option (C)