Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$
Hints are appreciated. Thanks in advance.
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2$\begingroup$ Voted to reopen. I sort of see the point of there not being enough context and detail but given the (extremely) good answer/s it just seems wrong to keep it closed. $\endgroup$– DRFCommented May 12, 2017 at 20:44
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3 Answers
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As $x\ne0,$ on simplification we find $$x^2+1=\sqrt3x$$
Squaring we get $$x^4+2x^2+1=3x^2\iff x^4-x^2+1=0$$
Now $$ x^6+1=(x^2+1)(x^4-x^2+1)$$
answered May 11, 2017 at 17:20
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$\begingroup$ You never fail to impress me. Thanks ! $\endgroup$– MathxxCommented May 11, 2017 at 17:32
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2$\begingroup$ You didn't complete the answer, but, I guess from where you left it, it's intuitively obvious. I guess, then, I'm stating the obvious that the answer is 0 $\endgroup$ Commented May 11, 2017 at 22:30
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$\begingroup$ From the last line of the answer, it seems like x is 0, but the first step of answer started off with the assumption that x is not equal to 0. $\endgroup$ Commented May 13, 2017 at 5:40
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1$\begingroup$ @ChiragBhatia-chirag64, From the last line $$x^6+1=0$$ not $x$ $\endgroup$ Commented May 13, 2017 at 5:41
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$\begingroup$ Got it, my mistake. $\endgroup$ Commented May 13, 2017 at 5:45
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Solving for $$x^2-\sqrt3x+1=0,$$
$$x=\dfrac{\sqrt3\pm i}2=\cos\dfrac\pi6\pm i\sin\dfrac\pi6=e^{\pm i\pi/6}$$
$$\implies x^6=e^{\pm i\pi}=-1$$
answered May 11, 2017 at 17:28
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1$\begingroup$ Thanks for the downvote! I hope I'll get a chance rectify my one mistake. $\endgroup$ Commented May 13, 2017 at 5:51
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Another way:
$$x^{18} + x^{24} = x^{21} (x^3 + \frac{1}{x^3})\\ = x^{21} ((x+\frac{1}{x})^3 - 3(x+\frac{1}{x}))\\ = x^{21}(3\sqrt{3} - 3\sqrt{3}) = 0. $$
answered May 13, 2017 at 5:48