# Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$

Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$

I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.

Hint: $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)=18.$$

Let $$x+\frac{1}{x}=t.$$ Then your equation reduces to $(t(t^2-3))=18$. Solve this cubic and you get $x+\frac{1}{x}.$

Now I think you can carry on after that.

• Did you try solving that cubic? Sep 14 '15 at 14:42
• That's the problem. The cubic seems to be impossible to solve. Sep 14 '15 at 14:43
• Not impossible, but surely there must be easier ways that trying $11$th power of each root $\left\{\frac{\sqrt[3]{81+\sqrt{6558}}}{3^{2/3}}+\frac{1}{3} \sqrt[3]{243-3 \sqrt{6558}}, -\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{81+\sqrt{6558}}}{2\ 3^{2/3}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{243-3 \sqrt{6558}},-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{81+\sqrt{6558}}}{2\ 3^{2/3}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{243-3 \sqrt{6558}}\right\}$ Sep 14 '15 at 14:46
• Do you have any suggestions Macavity? Sep 14 '15 at 14:51
• I was able to solve for one of the values of $t$ by trial and error. It's enough to solve the problem. Thanks. Sep 14 '15 at 14:58

In the equation $x^3+\frac{1}{x^3}=18$, multiply everything by $x^3$ to get $x^6-18x^3+1=0$. Then let $y=x^3$ and solve $y^2-18y+1=0$, then substitute back in to get $x$. Then you can input the value into $x^{11}+\frac{1}{x^{11}}$ directly.

It winds up being $x^{11}+\frac{1}{x^{11}}=39603$ (regardless of whether you use the plus or the minus in the quadratic formula).

• I already considered this, but it's computationally impractical, assuming computers are not allowed. Sep 14 '15 at 14:48
• Are there any bounds on what the values of $x^{11}+\frac{1}{x^{11}}$ will be? I mean to say is it a real number .
– user210387
Sep 14 '15 at 14:51
• @Rememberme Yes, it is a real number (so long as we avoid $x=0$ at least). Sep 14 '15 at 14:52
• Then we can just consider the real solution for the cubic
– user210387
Sep 14 '15 at 14:54
• @BenS. Then I suppose you took only the real cube root of $y_{\pm}$. Otherwise, you do get a complex value in addition to $39603$. Sep 14 '15 at 14:55

A little deus ex machina but this seems to be related to the golden ratio $\phi = \frac{1+\sqrt 5}{2}$, which is the solution of $x+\frac1x = 1$, and to the Fibonacci sequence $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657, \ldots$

In particular $x^3+\frac1{x^3} = 18$ has the solutions $x^3=5+8\phi$ and its reciprocal $-8\phi+13$, which requires $x=1+\phi$ or its reciprocal $-\phi+2$, and so $x^{11}=10946+17711\phi$ or its reciprocal $-17711\phi + 28657$ which gives $$x^{11} + \frac{1}{x^{11}} = 10946+17711\phi -17711\phi + 28657 = 39603.$$

• This is beautiful
– user210387
Sep 18 '15 at 9:29

If $x^3 + \frac{1}{x^3} = 18$ then squaring gives $x^6 + 2 + x^{-6} = 18^2$ so $x^6 + x^{-6} = 322$. Squaring again gives $x^{12} + x^{-12} = 322^2 - 2 = 103682$

Then $(x + x^{-1})(x^{11} + x^{-11}) = (x^{12} + x^{-12}) + (x^{10} + x^{-10})$.

Now to me it seems clear that since we have $$(x+ x^{-1})^3 = (x^3 + x^{-3}) + 3(x + x^{-1}),$$ if we perform the substitution $t = x+ x^{-1}$ as suggested by @Remember_Me, then we get $t^3 - 3t - 18 = 0$. Now, doing the rational root test yields the root $t=3$, and then we can solve for the other two roots, but all choices will give the same result for $x^{11} + x^{-11}$, so we may as well take $x+ x^{-1} = 3$.

Then (squaring each time) we have $x^2 + x^{-2} = 7$, $x^{4} + x^{-4} = 47$ and $x^{8} + x^{-8} = 2207$ and so $(x^8 + x^{-8})(x^2 + x^{-2}) = (x^{10} + x^{-10}) + (x^6 + x^{-6}).$ Thus $$x^{10} + x^{-10} = 2207 \times 7 - 322 = 15127,$$ and so finally we get (from the above equation) $$3(x^{11} + x^{-11}) = 103682 + 15127 = 118809,$$ so $x^11 + x^{-11} = 39603$, and we avoided...most...of the messy calculation!

Our old friend the golden ratio, or rather its enhancement by $1$, namely $a=1+\phi$, along with its reciprocal, is lurking behind this problem.

Let $a$ and $\bar a$ be respectively the larger and smaller root of $x^2-3x+1=0$. We have $a\bar a=1$ and $a+\bar a=3$. It's easy to show that $a^3+\bar a^3=18$. Also,$$a^{11}+\bar a^{11}=(a^5+\bar a^5)(a^6+\bar a^6)-(a+\bar a).$$After breaking $a^5+\bar a^5$ down to $(a^3+\bar a^3)[(a+\bar a)^2-2]-(a+\bar a)$, and $a^6+\bar a^6$ to $(a^3+\bar a^3)^2-2$, we get the answer$$a^{11}+\bar a^{11}=(18\times7-3)(18^2-2)-3=39,\!603.$$(Here we are ignoring the complex roots.)