Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$ [closed]

Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$

Hints are appreciated. Thanks in advance.

• 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.
– DRF
Commented May 12, 2017 at 20:44

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 $$x^6+1=(x^2+1)(x^4-x^2+1)$$

• You never fail to impress me. Thanks ! Commented May 11, 2017 at 17:32
• 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 Commented May 11, 2017 at 22:30
• 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. Commented May 13, 2017 at 5:40
• @ChiragBhatia-chirag64, From the last line $$x^6+1=0$$ not $x$ Commented May 13, 2017 at 5:41
• Got it, my mistake. Commented May 13, 2017 at 5:45

Solving for $$x^2-\sqrt3x+1=0,$$

$$x=\dfrac{\sqrt3\pm i}2=\cos\dfrac\pi6\pm i\sin\dfrac\pi6=e^{\pm i\pi/6}$$

$$\implies x^6=e^{\pm i\pi}=-1$$

• Thanks for the downvote! I hope I'll get a chance rectify my one mistake. Commented May 13, 2017 at 5:51

Another way:
$$x^{18} + x^{24} = x^{21} (x^3 + \frac{1}{x^3})\\ = x^{21} ((x+\frac{1}{x})^3 - 3(x+\frac{1}{x}))\\ = x^{21}(3\sqrt{3} - 3\sqrt{3}) = 0.$$