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,
$x^3+\frac{1}{x^3}+3\sqrt3=3\sqrt3$
⇒ $x^3+\frac{1}{x^3}=3\sqrt3-3\sqrt3$
⇒ $x^3+\frac{1}{x^3}=0$
⇒ $x^6+1=0$
⇒ $x^6=–1$
Substitute the value of $x^6=–1$ in given expression,
$x^{72}+x^{66}+x^{54}+x^{24}+x^6+1$
$=(x^6)^{12}+(x^6)^{11}+(x^6)^{9}+(x^6)^{4}+x^6+1$
$=(–1)^{12}+(–1)^{11}+(–1)^{9}+(–1)^{4}+(–1)+1$
$=1-1-1+1-1+1=0$
Hence, the correct answer is $0$.