An algebra problem by Koushik Vennelakanti

$\begin{array}{rlll} x + \dfrac{1}{x} &= &\sqrt{3} &\small \text{[Cube both sides]} \\ \\ x^3 + 3x^2 \left( \dfrac{1}{x} \right) + 3x \left( \dfrac{1}{x^2} \right) +\dfrac{1}{x^3} &= &\left( \sqrt{3} \right)^3 \\ \\ x^3 + 3 \left(x + \dfrac{1}{x} \right) + \dfrac{1}{x^3} &= &3 \sqrt{3} \qquad &\small \text{[Substitute in original equation]} \\ \\ x^3 + 3 \sqrt{3}+ \dfrac{1}{x^3} &= &3 \sqrt{3} \\ \\ x^3 + \dfrac{1}{x^3} &= &0 &\small \text{[Since } x \neq 0, \text{ multiply by } x^3] \\ \\ x^6 + 1 &= &0 \\ \\ x^6 + 2 &= &\boxed{1} \end{array}$

Square both sides you get $x^2+ \frac{1}{x^2}=1$ then let $m=x^2$ Then $m+ \frac{1}{m}=1$ $m^2+ 1=m$ Both sides times m $m^2-m+ 1=0$ Both sides times (m+1) we get $m^3+1=0$ then $m^3=-1$ because $m=x^2$ then $x^6=-1$ $x^6+2=1$

x+1/x=3^(1/2)

(x+1/x)^2=(3^(1/2))^2

x^2+2+1/(x^2)=3

x^2+1/(x^2)=1

x^4+1=x^2 .............1

x^6+x^2=x^4 ..........2 +

x^6+x^4+x^2+1=x^2+x^4

x^6+1=0

x^6+2=1

So, ans : 1

$\begin{aligned} x + \frac 1x & = \sqrt 3 & \small \color{#3D99F6} \text{Rearranging} \\ \implies x & = \sqrt 3 - \frac 1x & \small \color{#3D99F6} \text{Multiplying both sides by }x \\ x^2 & = \sqrt 3x - 1 & \small \color{#3D99F6} \text{Multiplying both sides by }x \\ x^3 & = \sqrt 3x^2 - x & \small \color{#3D99F6} \text{Note that }x^2 = \sqrt 3x - 1 \\ & = \sqrt 3(\sqrt 3x - 1) - x \\ \implies x^3 & = 2x - \sqrt 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^6 & = 4x^2 - 4\sqrt 3 x + 3 \\ & = 4(\sqrt 3x - 1) - 4\sqrt 3 x + 3 \\ & = -1 \\ \implies x^6 + 2 & = - 1 + 2 = \boxed{1} \end{aligned}$