Question : If $(x+\frac{1}{x})^2=3$, then the value of $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$ is:

Option 1: $0$

Option 2: $1$

Option 3: $84$

Option 4: $206$

Correct Answer: $0$

Solution : Given: $(x+\frac{1}{x})^2=3$
We know that the algebraic identity is $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$.
$(x+\frac{1}{x})^2=3$
By taking the square root on both sides of the above equation, we get,
$(x+\frac{1}{x})=\sqrt3$
By taking the cube on both sides of the above equation, we get,
$(x+\frac{1}{x})^3=(\sqrt3)^3$
$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=3\sqrt3$
Substitute the value of $(x+\frac{1}{x})=\sqrt3$ in above equation,
$x^3+\frac{1}{x^3}+3\sqrt3=3\sqrt3$
⇒ $x^3+\frac{1}{x^3}=3\sqrt3–3\sqrt3$
⇒ $x^3+\frac{1}{x^3}=0$
⇒ $x^6+1=0$
Substitute the value of $x^6+1=0$ in the given expression,
$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$
$=x^{200}(x^6+1)+x^{84}(x^6+1)+x^{12}(x^6+1)+x^6+1$
$=x^{200}\times0+x^{84}\times0+x^{12}\times0+0=0$
Hence, the correct answer is $0$.