Question

Let $\alpha ,\beta ,\gamma$ be the roots of the equation $x^3-3x^2+1=0$ then find the value of ${\alpha }^2+{\beta }^2+{\gamma }^2$.

Answer 1

Given, $\alpha ,\beta ,\gamma$ be the roots of the equation $x^3-3x^2+1=0$

Then, from the relation between roots and coefficients we get,

$\alpha +\beta +\gamma =3$ ………..$\left(1\right)$

$\alpha \beta +\beta \gamma +\gamma \alpha =0$ ………$\left(2\right)$

$\alpha \beta \gamma =-1$ ……….$\left(3\right)$

Now,

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

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

$=3^2-2.0=9$ [using $\left(1\right)$ & $\left(2\right)$].