Question : If $a+\frac{1}{a}=\sqrt{3}$, then the value of $a^{18}+a^{12}+a^{6}+1$ is:
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: 4
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Correct Answer: 0
Solution :
Given: $a^{18}+a^{12}+a^{6}+1$
= $a^{6}(a^{12}+1)+1(a^{12}+1)$
= $(a^{6}+1)(a^{12}+1)$-----------------------------------------------(i)
Now, $a+\frac{1}{a} = \sqrt{3}$
Squaring both sides,
$(a+\frac{1}{a})^2 = (\sqrt{3})^2$
⇒ $(a)^2+(\frac{1}{a})^2+2×a×\frac{1}{a} = 3$
⇒ $(a)^2+(\frac{1}{a})^2 = 3-2 = 1$
Cubing both sides,
$((a)^2+(\frac{1}{a})^2)^3 = 1^3$
⇒ $(a)^6+(\frac{1}{a})^6+3×((a)^2+(\frac{1}{a})^2) = 1^3$
⇒ $(a)^6+(\frac{1}{a})^6 = 1 - 3$
⇒ $\frac{((a)^{12}+1)}{(a)^6} = -2$
⇒ $a^{12}+1+2a^6 = 0$
⇒ $(a^{6}+1)^2 = 0$
⇒ $(a^{6}+1) = 0$ -------------------------------------------------(ii)
Substituting (ii) in (i)
$(a^{6}+1)(a^{12}+1)$ = 0
Thus, $a^{18}+a^{12}+a^{6}+1$ = 0
Hence, the correct answer is 0.
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