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. $18$
• B. $10$
• C. $15$
• D. $36$

Answer 1

The correct option is A $18$

${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-1$
And ${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha )\Rightarrow \alpha \beta \gamma =-2$
Then
${({\alpha }^2+{\beta }^2+{\gamma }^2)}^2=\sum {\alpha }^4+2\sum {\beta }^2{\gamma }^2=\sum {\alpha }^4+2({(\sum \beta \gamma )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma ))\Rightarrow {\alpha }^4+{\beta }^4+{\gamma }^4=36-2({(-1)}^2-2(-2)2)=18$