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If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$

The question would have been simple if it asked us to prove the other way round.

We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take 1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means?

NOTE: This problem is from a textbook of mine.

rah4927
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  • The title isn't supposed to be the first line of your question. The statement in the title is false since there are two real such $x$s (and even more complex solutions). – Git Gud Nov 27 '13 at 11:22
  • The other real root is related: $x=\sqrt{3}-\sqrt{2}$ is also a solution. – coffeemath Nov 27 '13 at 11:24
  • @rahul Another user took the liberty of partially edting the mathjax in your question and your last edit overrode it. I rolledback to the said user's edit. If you wish to edit something, please edit the current version. – Git Gud Nov 27 '13 at 11:26
  • @Git Gud,I think we can leave out the complex solutions as the book doesn't cover them. – rah4927 Nov 27 '13 at 11:26
  • Which book is that? – lhf Nov 27 '13 at 11:37

4 Answers4

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$$t+\frac1t=18\sqrt3\iff t^2-(2\cdot9\sqrt3)t+1=0\iff t_{1,2}=\frac{9\sqrt3\pm\sqrt{(9\sqrt3)^2-1\cdot1}}1=$$

$$=9\sqrt3\pm\sqrt{81\cdot3-1}\quad=\quad9\sqrt3\pm\sqrt{243-1}\quad=9\sqrt3\pm\sqrt{242}\quad=\quad9\sqrt3\pm\sqrt{2\cdot121}=$$

$$=9\sqrt3\pm\sqrt{2\cdot11^2}\quad=\quad9\sqrt3\pm11\sqrt2\quad\iff\quad x_{1,2}^3=9\sqrt3\pm11\sqrt2=(a\sqrt3+b\sqrt2)^3=$$

$$=(a\sqrt3)^3+(b\sqrt2)^3+3(a\sqrt3)^2b\sqrt2+3a\sqrt3(b\sqrt2)^2\ =\ 3a^3\sqrt3+2b^3\sqrt2+9a^2b\sqrt2+6ab^2\sqrt3$$

$$\iff3a^3+6ab^2=9=3+6\quad,\quad2b^3+9a^2b=\pm11=\pm2\pm9\iff a=1,\quad b=\pm1.$$

Lucian
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  • +1, I did the same... your solution have less steps which is nice to see ... :) –  Nov 27 '13 at 12:09
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Set $a=x+\frac{1}{x}, b\sqrt3=a$. Then we want that $$a^3-3a=18\sqrt3\iff b^3-b=6\iff(b-2)((b+1)^2+2)=0\iff b=2$$ So $$2\sqrt3=x+\frac{1}{x}\iff x^2-2\sqrt{3}x+1=0\iff (x-\sqrt3)^2-2=0 \iff x=\sqrt3\pm\sqrt2$$

chubakueno
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$x^3+x^{-3}=18\sqrt{3}\Rightarrow x^6+1=18\sqrt{3}x^3\Rightarrow x^6-18\sqrt{3}x^3+1=0$

From there if you look at it like a quadratic equation, you can find 2 solutions for $x^3$ So x would simply be the cube roots of that. and $x^3=a$ has only 1 real root for any real-valued a.

wfw
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    I'm not sure what you're saying, but it's obvious from looking at the equation is $α$ is a solution, then so $α^{−1}$ is, so claiming it has only one solution is wrong. – Git Gud Nov 27 '13 at 11:35
  • It $x^3=a$ has a unique solution for the two $a$ that you find as a solution of $a^2-18\sqrt{3}a+1=0$ so in fact you get a solution for every $a$. In the end you have two solutions. – Beni Bogosel Nov 27 '13 at 11:52
  • @GitGud $x^3=a$ has only 1 real solution. We come up with two solutions for $y^2-18\sqrt{3}y+1=0$ and then only one corresponding x for either y such that $x^3=y$ – wfw Nov 27 '13 at 11:55
  • @GitGud, expand your comment that only if $\alpha = 1$ do we have $\alpha = \alpha^{-1}$, and $x = 1$ certainly isn't a solution. So there are an even number of real solutions. Neat. – vonbrand Feb 10 '14 at 03:18
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$x^3+\frac{1}{x^3}=18\sqrt{3}\Rightarrow x^6-18\sqrt{3}x^3+1=0 $

we asssume $y=x^3$

$y^2-18\sqrt{3}y+1=0\Rightarrow y=\frac{18\sqrt{3}\pm\sqrt{968}}{2}=\frac{18\sqrt{3}\pm22\sqrt{2}}{2}=9\sqrt{3}\pm11\sqrt{2}$

$x^3-(9\sqrt{3}\pm11\sqrt{2})=0$

let $x=a\sqrt{3}+b\sqrt{2}$ now,

$x^3=3\sqrt{3}a^3+2\sqrt{2}b^3+3.3a^2b\sqrt{2}+3.a\sqrt{3}.2b^2=(3a^3+6ab^2)\sqrt{3}+(2b^3+9a^2b)\sqrt{2}$

Now, $3a^3+6ab^2=9$ and $2b^3+9a^2b=11$

i.e., $a^3+2ab^2=3$ and $2b^3+9a^2b=11$.

Now,

for $a^3+2ab^2=3$ one possible case would be $a^3=1,ab^2=1$ (I am not saying this is "the" one)

i.e., $a=1 ,b=\pm 1$

Now,

for $2b^3+9a^2b=11$ one possible case would be $b^3=1,a^2b=1$ (I am not saying this is "the" one)

i.e., $b=1, a=\pm 1$

but then we need both $a^3+2ab^2=3$ and $2b^3+9a^2b=11$ to satisfy at once.

So, only possibility would be $a=1,b=1$ which would imply $x=\sqrt{3}+\sqrt{2}$

Note : I took only $+11$ in $9\sqrt{3}\pm11\sqrt{2}$ , I would leave it to you to do the same for $-11$ case.

Good Luck!