Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+x^2+x+1=0$, then ${\alpha }^4+{\beta }^4+{\gamma }^4$, is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+x^2+x+1=0$
We have
$s_1=\alpha +\beta +\gamma =-1s_2=\alpha \beta +\beta \gamma +\alpha \gamma =1s_3=\alpha \beta \gamma =-1$
Now ${\alpha }^4+{\beta }^4+{\gamma }^4={({\alpha }^2+{\beta }^2+{\gamma }^2)}^2-2({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)={({(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\alpha \gamma ))}^2-2({(\alpha \beta +\beta \gamma +\alpha \gamma )}^2-2(\alpha +\beta +\gamma )\left(\alpha \beta \gamma \right))={({(-1)}^2-2\left(1\right))}^2-2({\left(1\right)}^2-2(-1)(-1))=1+2=3$