Question

Given that $\alpha ,\beta ,\gamma$ are the roots of cubic equation $x^3-3x^2+2x+\frac{2}{3}=0$ the value of ${\alpha }^4+{\beta }^4+{\gamma }^4$ is equal to

Answer 1

$x^3-3x^2+2x+\frac{2}{3}=0rootsare\alpha ,\beta ,\gamma \alpha +\beta +\gamma =3\alpha \beta +\beta \gamma +\gamma \alpha =2\alpha \beta \gamma =\frac{-2}{3}{\alpha }^2+{\beta }^2+{\gamma }^2=({\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )=9-4=5{\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2={(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )={\left(2\right)}^2+2.\frac{2}{3}.3{\alpha }^4+{\beta }^4+{\gamma }^4={({\alpha }^2+{\beta }^2+{\gamma }^2)}^2-2({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)=25-16=9$