Question

If $\alpha +\beta +\gamma =6,{\alpha }^2+{\beta }^2+{\gamma }^2=14$ and ${\alpha }^3+{\beta }^3+{\gamma }^3=36$, then ${\alpha }^4+{\beta }^4+{\gamma }^4=$

• A. $98$
• B. $103$
• C. $24$
• D. $342$

Answer 1

The correct option is A $98$

${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )\Rightarrow 6^2=14+2(\alpha \beta +\beta \gamma +\gamma \alpha )\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =11$

${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha )\Rightarrow 36-3\alpha \beta \gamma =\left(6\right)(14-11)\Rightarrow \alpha \beta \gamma =6$

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

$\Rightarrow {\left(14\right)}^2=\sum {\alpha }^4+2({\left(11\right)}^2-2\left(6\right)\left(6\right))\Rightarrow {\alpha }^4+{\beta }^4+{\gamma }^4=169-2(121-72)=98$