Question

The sum of the cubes of all the roots of the equation $x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$ is _______

Answer 1

$x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$
$x=0$ is not the root of this equation so divide it by $x^{2}$
$x^{2}-3 x-2+\frac{3}{x}+\frac{1}{x^{2}}=0$
$x^{2}+\frac{1}{x^{2}}-2+2-3\left(x-\frac{1}{x}\right)-2=0$
$\left(x-\frac{1}{x}\right)^{2}-3\left(x-\frac{1}{x}\right)=0$
$x-\frac{1}{x}=0$
$x^{2}-1=0$
$x=\pm 1$
$\alpha=1, \beta=-1$
$x-\frac{1}{x}=3$
$x^{2}-3 x-1=0$
$\gamma+\delta=3$
$\gamma \delta=-1$
$1-1+(\gamma+\delta)\left((\gamma+\delta)^{2}-3 \gamma \delta\right)$
$0+3(9-3(-1))$
$+3(12)=36$