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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Question : If $(a+\frac{1}{a})^{2}=3$, then the value of $(a^{6}-\frac{1}{a^{6}})$ will be:
Question : If $x+\frac{1}{x}=\sqrt{3}$, the value of $(x^{18}+x^{12}+x^{6}+1)$ is:
Question : If $\frac{1}{a}(a^2+1)=3$, then the value of $\frac{a^6+1}{a^3}$ is:
Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Question : If $x+\frac{1}{x}=6$, then find the value of $\frac{3 x}{2 x^2-5 x+2}$.
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