Question

Find the polynomial equation whose roots are the squares of the roots of $x^3-x^2+8x-6=0$

Answer 1

Let b$\alpha ,\beta ,\gamma$ are the roots of $x^3-x^2+8x-6=0$

$\alpha +\beta +\gamma =1$

$\alpha \beta +\beta \gamma +\gamma \alpha =8$

$\alpha \beta \gamma =6$

Let, $P={\alpha }^2+{\beta }^2+{\gamma }^2$

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

$=1-2\left(8\right)=-15$

Let, $Q={\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2$

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

$=(8{)}^2-2\left(\alpha \beta \gamma \times \alpha +\alpha \beta \gamma \times \beta +\alpha \beta \gamma \times \gamma \right)$

$=64-12\left(\alpha +\beta +\gamma \right)$

$=64-12\left(1\right)$

$=52$

Let, $R={\alpha }^2{\beta }^2{\gamma }^2$

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

$=6^2=36$

Equation of polynomial whose roots are ${\alpha }^2,{\beta }^2,{\gamma }^2$

$x^3-\left(P\right)x^2+Qx-R=0$

$x^3+15x^2+52x-36=0$