Question : If $x+\frac{1}{x}=\sqrt{3}$, the value of $(x^{18}+x^{12}+x^{6}+1)$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
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Correct Answer: 0
Solution :
Given: $x+\frac{1}{x}=\sqrt{3}$ (equation 1)
We know that the algebraic identity is $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$
$x+\frac{1}{x}=\sqrt{3}$
Take the cube on both sides of the above equation, we get,
$(x+\frac{1}{x})^3=(\sqrt{3})^3$
$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=3\sqrt{3}$
Substitute the value from equation 1 in the above equation and we get,
$x^3+\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$
$x^3+\frac{1}{x^3}+3\sqrt{3}–3\sqrt{3}=0$
$x^6+1=0$
$x^6=–1$
The expression $(x^{18}+x^{12}+x^{6}+1)$ can be written as $(x^6)^3+(x^6)^2+x^{6}+1$
Substitute the value of $x^6=–1$ in above equation, we get,
$=(–1)^3+(–1)^2–1+1$
$=–1+1 – 1+1$
$=0$
Hence, the correct answer is 0.
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