# Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$

If $$x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$$, find the value of $$x-\dfrac{1}{x}$$.

Here's what I've tried so far.

\begin{align} \left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\ \rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0 \end{align}

Got stuck here since I didn't know how to solve this cubic equation.

I also tried factorizing $$x^3-\dfrac{1}{x^3}$$.

\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\ &= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\ &= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\ &= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right) \end{align}

Again, I didn't know what I could do with this.

• Hint. $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)^3-3xy(x-y)$.
– RDK
Commented Oct 18, 2022 at 15:42
• You should be able to figure this out with the hints. However, MSE expects that you show your own efforts. This is not a portal to dump questions. Commented Oct 18, 2022 at 15:44
• Alright, you may think this is not worth doing because of $xy$. Then, try to "eliminate" $xy$. What's $y$ which makes $xy=1$? (With a bit of calculus, you can easily know the critical point, but with precalculus, just try solving this.)
– RDK
Commented Oct 18, 2022 at 16:05
• Try making the variable substitution $u = x^3$, solve for $u$, and take the cube root.
– Dan
Commented Oct 18, 2022 at 16:18
• Cubic Formula with $ax^3+cx+d=0$: $x=\sqrt[3]{\left( - \frac d {2a} \right)+\sqrt{\frac{d^2}{4a^2}+\frac{c^3}{27a^3}}} + \sqrt[3]{\left( -\frac {d}{2a} \right)-\sqrt{\frac{d^2}{4a^2}+\frac{c^3}{27a^3}}}$
– RDK
Commented Oct 18, 2022 at 22:20

Solve for $$x$$ first. You can use the quadratic formula.

$$x^3 - \frac1{x^3} = 108 + 76\sqrt2 \implies x^6 - (108+76\sqrt2)x^3 - 1 = 0$$

$$\implies x^3 = \frac{108+76\sqrt2 \pm \sqrt{(108+76\sqrt2)^2 + 4}}2 = 54 + 38\sqrt2 + 3\sqrt{645 + 456 \sqrt2}$$

$$\implies x = \sqrt[3]{54 + 38\sqrt2 + 3\sqrt{645 + 456 \sqrt2}} = \frac{3+2\sqrt2}2 + \frac12 \sqrt{21 + 12\sqrt2}$$

The hardest part is de-nesting the cube root (see below). Otherwise take WA at its word.

It follows that

$$\frac1x = \frac2{3 + 2\sqrt2 + \sqrt{21 + 12\sqrt2}} = -\frac{3+2\sqrt2}2 + \frac12 \sqrt{21 + 12\sqrt2}$$

and you can easily find $$x-\frac1x$$ from here.

De-nesting the cube root

Observe that

$$645 + 456 \sqrt2 = (21 + 12\sqrt2) (17 + 12\sqrt2) = (21 + 12\sqrt2) (3 + 2\sqrt2)^2 \\ \implies x^3 = 54 + 38\sqrt2 + 9\sqrt{21+12\sqrt2} + 6\sqrt2\sqrt{21+12\sqrt2}$$

Suppose we can decompose the cube root into the form

$$x = a + b \sqrt2 + c \sqrt{21+12\sqrt2}$$

where $$a,b,c\in\Bbb Q$$.

Taking cubes on both sides and matching up coefficients, we get the system of equations

$$\begin{cases} a^3 + 6 a b^2 + 63 a c^2 + 72 b c^2 = 54 \\ 2b^3 + 3 a^2 b + 36 a c^2 + 63 b c^2 = 38 \\ 7c^3 + a^2 c + 2 b^2 c = 3 \\ 2c^3 + a b c = 1 \end{cases}$$

Solving the last equation for $$c^2 = 3ab-a^2-2b^2$$ and substituting that into the first two equations yields yet another system,

$$\begin{cases} -62a^3 + 117a^2b + 96ab^2 - 144b^3 = 54 \\ -36a^3 + 48a^2b + 117 ab^2 - 124b^3 = 38 \end{cases}$$

Since both polynomials are homogeneous with degree $$3$$, suppose that $$a,b$$ are proportional. Let $$b=ka$$ where $$k\in\Bbb Q$$, so we have

$$\begin{cases} a^3(-62 + 117k + 96k^2 - 144k^3) = 54 \\ a^3(-36 + 48k + 117k^2 - 124k^3) = 38 \end{cases}$$

Eliminate $$a^3$$ by division to get a cubic equation in $$k$$.

$$\frac{a^3(-62 + 117k + 96k^2 - 144k^3)}{a^3(-36 + 48k + 117k^2 - 124k^3)} = \frac{27}{19} \implies 612k^3 - 1335k^2 + 927k - 206 = 0$$

Now apply the rational root theorem. If you start with the smallest divisors of $$612$$ and $$206$$, you'll soon find $$k=\frac23$$ (and we can show this is the only rational solution for $$k$$). It follows that

$$b=\frac23a \implies 16a^3 = 54 \implies a = \frac32 \implies b = 1 \implies c = \frac12$$

• Wow. You did very difficult computations. Commented Oct 18, 2022 at 17:26
• Thank you, but I wouldn’t be able to decompose that cube root without WA! Commented Oct 19, 2022 at 7:36
• In the first part of ‘decomposing the cube root’, how did you manage to make that observation? Or was it just pure luck/intuition Commented Oct 20, 2022 at 7:21
• The second half of this answer is more an attempt at explaining how WA is getting its result. One could instead assume a decomposition of the form $a+b\sqrt2+c\sqrt{645+456\sqrt2}+d\sqrt2\sqrt{645+456\sqrt2}$ but then the system of equations that follows is a bit more complicated (again with a single solution over the rationals, namely $(a,b,c,d)=\left(\frac32,1,\frac32,-1\right)$). Commented Oct 20, 2022 at 16:03
• @user170231 why would you have the ‘d’ term when the original cube root doesn’t have it? Commented Oct 23, 2022 at 6:48

\begin{align} \newcommand{w}{\omega} & x^3-\frac{1}{x^3}=108+76\sqrt{2}. \\ & 108+76\sqrt{2} = \left(x-\frac 1 x\right)\left(x^2+1+\frac 1 {x^2}\right) \\ & = \left(x-\frac 1 x \right)^3+3x\cdot \frac 1 x \left(x-\frac 1 x\right). \\ \ \\ & u^3+3u=108+76\sqrt{2}. \\ & 108+76\sqrt{2}= 4(27+19\sqrt{2})=4(5+3\sqrt{2})(1+\sqrt{2})^2. \\ & (5+3\sqrt{2})^2+3=46+30\sqrt{2} \neq 12+8\sqrt{2}. \\ & \vdots \\ & (3+2\sqrt{2})^2+3=20+12\sqrt{2}=4(5+3\sqrt{2}). \\ \therefore \; & u=3+2\sqrt{2}. \end{align}

Substituting $$x-\dfrac 1 x=3+2\sqrt{2}$$, it works.

• Wow, that’s amazing. Any tips on how to factor $a + b\sqrt{c}$, like you did here? Commented Oct 19, 2022 at 7:35
• How would you know that you need to check integers for the rational an irrational part? Commented Oct 19, 2022 at 18:34

Note that

$$(x - \frac{1}{x})^3 = x^3 - 3x + \frac{3}{x} - \frac{1}{x^3}$$ $$(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}{x})$$ $$x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x})$$

So if you let $$t = x - \frac{1}{x}$$ (the value that you're ultimately planning to solve for), then your equation becomes:

$$t^3 + 3t = 108 + 76\sqrt{2}$$

This is a “depressed” cubic equation that you could solve with Cardano's Formula, but that gives you some ugly nested radicals. So instead, I'm going another route, by finding an equivalent polynomial equation with integer coefficients.

$$(t^3 + 3t - 108)^2 = (76\sqrt{2})^2$$ $$t^6 + 3t^4 - 108t^3 + 3t^4 + 9t^2 - 324t - 108t^3 - 324t + 11664 = 11552$$ $$t^6 + 6t^4 - 216t^3 + 9t^2 - 648t + 112 = 0$$

There's no general formula for a sixth-degree polynomial. And the Rational Root Theorem fails to provide us with any linear factors. But it turns out that a quadratic factor exists.

$$(t^2 - 6 t + 1) (t^4 + 6 t^3 + 41 t^2 + 24 t + 112) = 0$$

Solving the quadratic gives $$t = 3 \pm 2\sqrt{2}$$. Solving the remaining quartic gives 4 complex solutions:

$$t = -\frac{3}{2} + \sqrt{2} \pm \frac{3}{2} i \sqrt{7 - 4 \sqrt{2}}$$ $$t = -\frac{3}{2} - \sqrt{2} \pm \frac{3}{2} i \sqrt{7 + 4 \sqrt{2}}$$

But of these 6 roots, it turns out that only one of them is a valid solution to the original equation.

$$t = \boxed{3 + 2\sqrt{2}}$$

• nice! Although it's far from obvious to spot the quadratic factor. Commented Nov 23, 2022 at 14:10

Here is a suggestion to get an answer systematically - though not fully analytic.

Others have already noted that we need to solve for the desired $$u = x- \frac{1}{x}$$ in $$f(u) = u^3+3u- 108- 76\sqrt{2} = 0$$ Now observe that any expression of the form $$y = a + b \sqrt{2}$$, when taken to an integer power, reproduces this form, i.e. $$y^n = a_n + b_n \sqrt{2}$$. Since the constant part in $$f(u)$$ is also of this form, we can write $$u = a + b \sqrt{2}$$, which gives $$f(u) = a^3 + 6 a b^2 + 3 a -108 + \sqrt{2} (3 a^2 b +2 b^3 + 3 b - 76) = 0$$ This enables us to solve for the rational and the irrational parts separately, i.e. $$a^3 + 6 a b^2 + 3 a -108 = 0\\ 3 a^2 b +2 b^3 + 3 b - 76 = 0$$ where $$a$$ and $$b$$ are rational numbers. Note that it is not clear beforehand whether $$a$$ and $$b$$ are integers, as other answers have tested some integer combinations.

Now this is again two coupled third-degree equations so one might ask what has been won (maybe one of you has a nice idea how to solve this analytically).

I introduce now a way to manipulate these equations to come up with an iterative procedure with works without any knowledge or presupposition of the nature of $$a$$ and $$b$$. Also, it does not need any roots (of quadratic or cubic equations) which have been used earlier but which can become cumbersome.

Writing equivalently $$a^2 + 6 b^2 + 3 -\frac{108}{a} = 0\\ 3 a^2 +2 b^2 + 3 - \frac{76}{b} = 0$$ allows to isolate the terms with $$a^2$$ and $$b^2$$ by weighted adding, which gives

$$4 a^2 + 54/a + 3 = 114/b \\ 8 b^2 + 38/b + 3 = 162/a$$ Now let $$A = 1/a$$ and $$B = 1/b$$ to end with $$B = \frac{9 A}{19} + \frac{1}{38} + \frac{2}{57 A^2} \\ A = \frac{19 B}{81} + \frac{1}{54} + \frac{4}{81 B^2}$$ One can make this an iterative pair of equations; start with some $$A$$, obtain directly $$B$$ from the first equation, plug into the second and obtain a new $$A$$, repeat ...

Both equations are of the same functional form. Since there must be a solution, the first one should fall faster with $$A$$ then the second one with $$B$$, so this iteration should converge. Here are some values which are obtained when starting with $$A=1$$:

$$A=1 , B = 0.535, A = 0.3165, B = 0.526, A = 0.320, B = 0.520, A = 0.323, \cdots$$

Convergence is not very fast, but letting this go for a while leads to values which are approximating $$A=\frac13 , B = \frac12$$ and indeed one can check that this solves the equations.

So the result is $$u = 3 + 2 \sqrt{2}$$, q.e.d. $$\qquad \Box$$

• Hey, at the beginning of your answer, you mention that you need to solve for $u=x+\frac{1}{x}$, but that cubic equation only works if u is the conjugate of what you’ve assumed it to be. Commented Oct 20, 2022 at 7:36
• @avighnac Sorry, writing error -> corrected. Thank you for mentioning! Commented Oct 20, 2022 at 10:13

$$u^3+3u=108+76\sqrt{2}$$

A few posts here assumed $$u = a + b\sqrt{2}$$, then search for rational $$(a, b)$$

Let "slope", $$\displaystyle \;m_1 = b/a$$

$$\displaystyle u^2 = (a^2+2b^2) \;+\; (2ab)\,\sqrt{2}\qquad\; →m_2 = \frac{2\,m_1}{1+2\,m_1^2}$$

Note that $$m_2$$ has same sign as $$\displaystyle m_1 \quad →m_∞ = \frac{sgn(m1)}{\sqrt{2}}$$

Slope of RHS = $$\displaystyle \frac{76}{108} ≈ 0.7037 \;<\; \frac{1}{\sqrt{2}} ≈ 0.7071$$

$$u^3 ≈ RHS,\;m_1$$ likely convergent of RHS slope.

C:\>spigot -C 76/108
0/1
1/1
2/3
5/7
7/10
19/27


Convergent 2/3 work! Upon checking, $$(b,a) = (2,3)$$, not just the ratio.

$$u^3 + 3u = (99+70\sqrt{2}) + 3×(3+2\sqrt{2}) = 108+76\sqrt{2}$$

• (108+76√2)^(1/3) ≈ 6.00, which is already close to (3+2√2) ≈ 5.83. Commented Jan 16, 2023 at 14:06
• cbrt(x+y√D) = a+b√D, see math.stackexchange.com/a/4621416 Commented Jan 19, 2023 at 4:23