Question

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

Answer 1

Answer: 36

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