Question : If $(a+\frac{1}{a})^{2}=3$, then the value of $a^{18}+a^{12}+a^{6}+1$ is:
Option 1: 3
Option 2: 1
Option 3: 0
Option 4: 2
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Correct Answer: 0
Solution :
Given: $(a+\frac{1}{a})^{2}=3$
⇒ $(a+\frac{1}{a})=\sqrt3$
Cubing both sides,
$(a^3+\frac{1}{a^3})= (\sqrt3)^3-3\times\sqrt3$
⇒ $(a^3+\frac{1}{a^3})= 0$
Now,
$a^{18}+a^{12}+a^{6}+1$
$=a^{15}(a^3+\frac{1}{a^3})+a^{3}(a^3+\frac{1}{a^3})$
$=(a^{15}+a^3)(a^3+\frac{1}{a^3})$
$= 0$
Hence, the correct answer is 0.
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