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

Question

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.

Answer 1

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.

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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

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

Answer 2

I thought it might be helpful with answering now.

\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.

Answer 3

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.

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

Answer 4

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. $\qquad \Box$

Answer 5

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

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.

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