JEE Main 2022 — Mathematics Algebra

$${x}^{4}-3{x}^{3}-2{x}^{2}+3x+1=0$$x4−3x3−2x2+3x+1=0

Dividing by ${x}^{2}$x2 as $x\neq 0$x=0

$${x}^{2}-3x-2+\frac{3}{x}+\frac{1}{{x}^{2}}=0$$x2−3x−2+x3​+x21​=0

$${x}^{2}+\frac{1}{{x}^{2}}-2+2-3(x-\frac{1}{x})-2=0$$x2+x21​−2+2−3(x−x1​)−2=0

$${(x-\frac{1}{x})}^{2}-3(x-\frac{1}{x})=0$$(x−x1​)2−3(x−x1​)=0

$$\begin{matrix}x-\frac{1}{x}=0, & x-\frac{1}{x}=3\end{matrix}$$x−x1​=0,​x−x1​=3​

$$\begin{matrix}{x}^{2}-1=0, & {x}^{2}-3x-1=0\end{matrix}$$x2−1=0,​x2−3x−1=0​

$$x=\pm 1$$x=±1

For ${x}^{2}-3x-1=0$x2−3x−1=0 let the roots are \gamma &\delta

$$\begin{matrix}\alpha =1,\beta =-1,\gamma +\delta =3, & \gamma \delta =-1\end{matrix}$$α=1,β=−1,γ+δ=3,​γδ=−1​

$${\alpha }^{3}+{\beta }^{3}+{\gamma }^{3}+{\delta }^{3}$$α3+β3+γ3+δ3

$$=1-1+(\gamma +\delta )({(\gamma +\delta )}^{2}-3\gamma \delta )$$=1−1+(γ+δ)((γ+δ)2−3γδ)

$=3(9-3(-1))$=3(9−3(−1))$=36$=36