Question : If $x+\frac{1}{x}=\sqrt{3}$ then, the value of $x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$ is:
Option 1: $\sqrt{3}$
Option 2: $-\sqrt{3}$
Option 3: $1$
Option 4: $0$
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Correct Answer: $0$
Solution :
Given: $x+\frac{1}{x}=\sqrt{3}$
Cubing both sides, we get
$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=3\sqrt{3}$
⇒ $x^3+\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$
⇒ $x^6+1=0$
The expression is $x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$
$=x^{30}+x^{24}+x^{18}+x^{12}$
$=x^{30}+x^{24}+x^{12}(x^{6}+1)$
$=x^{30}+x^{24}$
$=x^{24}(x^{6}+1)$
$=0$
Hence, the correct answer is $0$.
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