Algebraically Solve $\left[a + b\sqrt{57}~\right]^3 = 540 + 84\sqrt{57}.$

Question

Unclear how valuable this posting is. It really should be limited to specifying that the goal is to denest one level of the radicals, in an expression like

$$\left[c + d\sqrt{D}\right]^{1/3} + \left[c - d\sqrt{D}\right]^{1/3} ~c,d,D \in \Bbb{Z}, ~D~ \text{is square free}.$$ As KCd indicated in a comment, following his answer, I totally overlooked that the rational root theorem is decisive for finding a rational value for the variable $a$.

So, the (remaining) problem is : what happens if

$$\left(a + b\sqrt{D}\right)^3 = c + d\sqrt{D},$$

where $a$ is irrational?

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$\underline{\text{The Problem}}$

I am looking for an Algebraic derivation that

$$\left[ ~a + b\sqrt{57} ~\right]^3 = \left[ ~540 + 84\sqrt{57} ~\right] ~: ~a,b \in \Bbb{R}$$

may be solved by $~(a,b) = (3,1).$

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$\underline{\text{My Background}}$

Some years ago, I survived self-studying :

• Calculus, Vol 1, 2nd Ed. (Tom Apostol, 1966)

• Through chapter 10, which includes Quadratic Reciprocity Law, of
  Elementary Number Theory (Uspensky and Heaslett, 1938)

• Chapters 1 and 2 only of
  An Introduction to Complex Function Theory (Bruce Palka, 1991).

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$\underline{\text{Problem Background}}$

I noticed a youtube problem: $~\displaystyle f(x) = x^3 + x - \frac{5}{8} = 0.$

Since I trial/error saw that $~f(1/2) = 0,$ I was able to use
polynomial long division to determine that the roots of $~f(x) = 0~$ are

$\displaystyle \left( ~\frac{1}{2}, \frac{-1 \pm i\sqrt{19}}{4} ~\right).$

As an exercise, I decided to practice using Cardano's Method against the equation:

$$x^3 + x - \frac{5}{8} = 0.$$

Setting

$$S + T = x, ~3ST = -1 \implies x^3 = S^3 + T^3 = S^3 + \left[\frac{-1}{3S}\right]^3 \implies $$

$$\left[S^3\right]^2 - \frac{5}{8}\left[S^3\right] - \frac{1}{27} = 0 \implies $$

$$S^3 = \frac{1}{2} ~\left[ ~\frac{5}{8} \pm \frac{7}{72}\sqrt{57} ~\right]$$

$$= \frac{1}{\left(12\right)^3} ~\left[540 \pm 84\sqrt{57}\right].$$

This implies that the equation

$$x^3 + x - \frac{5}{8} = 0$$

has the real root

$$\frac{1}{12} ~\left( ~\left[540 + 84\sqrt{57}\right]^{(1/3)} ~+~ ~\left[540 - 84\sqrt{57}\right]^{(1/3)} ~\right). \tag1 $$

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$\underline{\text{My Initial Work}}$

In order to simplify the expression in (1) above, I noted that

$$\left[a + b\sqrt{57}\right]^3 = \left[a^3 + 171ab^2\right] + \sqrt{57} ~\left[3a^2b + 57b^3\right].$$

So, I have the following two (non-linear) equations in two unknowns:

• Equation-1 : $~\displaystyle a^3 + 171ab^2 = 540.$

• Equation-2 : $~\displaystyle 3a^2b + 57b^3 = 84.$

Since I couldn't find an obvious line of attack to derive the $~(a,b) = (3,1)~$ solution to the above two equations, I took the preliminary step of verifying the solution. I used a somewhat convoluted method.

I reasoned that since the only real root of $~f(x) = x^3 + x - \frac{5}{8} = 0~$ is $~x = \frac{1}{2} = \left[\frac{3}{12} + \frac{3}{12}\right],~$ I must have that $~a = 3.~$ I was then able to verify that $~(a,b) = (3,1)~$ satisfied both of Equation-1 and Equation-2, above.

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$\underline{\text{My Subsequent Work}}$

Since the derivation process involves not knowing any of the actual roots to $~f(x) = 0,~$ the $~(a,b) = (3,1)~$ guesswork does not represent an analytical means of attack.

One approach is to substitute one value for another.

This leads to (for example)

$$3a^2 \left[ ~\frac{540 - a^3}{171a} ~\right]^{(1/2)} + 57\left[ ~\frac{540 - a^3}{171a} ~\right]^{(3/2)} = 84. \tag2 $$

Edit
The above expression does not represent a Gauss function.

My only other try is to try to use elementary Complex Analysis, by noting that

$$\left[a + ib\sqrt{57}\right]^3 = \left[a^3 - 171ab^2\right] + i\sqrt{57} ~\left[3a^2b - 57b^3\right].$$

If I could (somehow) obtain an appropriate expression of

$$(a + ib)^3 = \left[a^3 + 171ab^2\right] + i\sqrt{57} ~\left[3a^2b + 57b^3\right],$$

then, I could convert the RHS above into $~re^{i\theta},~$ thereby simplifying the cube root to

$$r^{1/3}e^{i[\theta + 2k\pi]/3} ~: ~k \in \{0,1,2\}.$$

However, I see no way of pursuing this last approach.

Answer 1

To do this in a non-ugly way, use the norm map $N(x+y\sqrt{57}) = x^2 - 57y^2$, which is multiplicative from $\mathbf Z[\sqrt{57}]$ to $\mathbf Z$.

Suppose $(a + b\sqrt{57})^3 = 540 + 84\sqrt{57}$ for some integers $a$ and $b$. Take the norm of both sides: $$ (a^2 - 57b^2)^3 = 540^2 - 57 \cdot 84^2 = -110592= (-48)^3, $$ which is equivalent to $a^2 - 57b^2 = -48$. Since $57$ and $48$ are both divisible by $3$, $a$ must be divisible by $3$, so write $a = 3c$. Plug that in and divide through by $3$ to get $$ 3c^2 - 19b^2 = -16, $$ or equivalently $3c^2 = 19b^2 - 16$. An example of an integral solution to that is obvious: $c = 1$ and $b = 1$, so we try $a = 3c = 3$ and $b = 1$ in the original equation and it works.

Answer 2

$(a+b\sqrt{D})^3 = (3ab^2D+a^3) + (b^3D+3a^2b)\sqrt{D} = x + y\sqrt{D}$

$x = (3ab^2D+a^3) \quad → \displaystyle b^2 = \frac{x-a^3}{3aD}$

$ y = b\,(Db^2+3a^2)$

$\displaystyle y^2 = \left(\frac{x-a^3}{3aD}\right)\left( \left(\frac{x-a^3}{3a}\right) + 3a^2 \right)^2$

Let $A=a^3$, and multiply both side by $27DA$, we have:

$27Dy^2A = (x-A)(x+8A)^2$

$64A^3 - 48x\,A^2 + (27Dy^2-15\,x^2)A - x^3 = 0$

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Derive similarly, we also have a cubic for $B = b^3$

$27B\,x^2 = (y-DB)\big(y + 8DB \big)^2$

$64D^3\,B^3 - 48D^2y\,B^2 + (-15Dy^2+27x^2)B - y^3 = 0$

Answer 3

My previous answer assumed $a,b \in \Bbb{R}$, solved algebraically for $\sqrt[3] {x+y\sqrt{D}} = a+b\sqrt{D}$
$(a^3,b^3)$ are real solutions of cubic equations, which may be messy.

If the goal is to simplify, and $a,b \in \Bbb{Q}$, we can do this numerically.

$\sqrt[3]{A ± \sqrt{R}} = a ± \sqrt{r} \quad ⇒ \quad a = \large\frac{\sqrt[3]{A+\sqrt{R}} \;+\; \sqrt[3]{A-\sqrt{R}}}{2}$

I discussed this very issue in HP Forum: HP50g simplifying a root
The link has Lua code to do simplifying (with confirmation) automatically.

lua> x, y, D = 540, 84, 57
lua> (cbrt(x + y*sqrt(D)) + cbrt(x - y*sqrt(D))) / 2
3.0000000000000004

lua> simp_cbrt4(x, y, D)
3       1       57

lua> simp_cbrt4(81, 30, -3) -- integer argument might produce halves
4.5     0.5     -3

$\displaystyle \sqrt[3]{540+84\sqrt{57}} = 3 + \sqrt{57}$

$\displaystyle \sqrt[3]{81+30\sqrt{-3}} = \frac{9+\sqrt{-3}}{2}$