Surd Conjugacy

Theorem

Let $F\subset\mathbb{R}$ be a ground field (think $F=\mathbb{Q}$).
Take $a,b,x,y\in F$ with $y\ge 0$ and $y$ not a square in $F$ (so $\sqrt y\notin F$).
Assume real radicals and that

$$\sqrt[3]{\,a+\sqrt b\,}=x+\sqrt y .$$

Then

$$\sqrt[3]{\,a-\sqrt b\,}=x-\sqrt y .$$

Proof

Now cube the assumed equality:

$$a+\sqrt b=(x+\sqrt y)^3 =(x^3+3xy)+(3x^2+y)\sqrt y .$$

By the lemma, comparing the $F$-part and the $\sqrt y$-part gives

$$a=x^3+3xy,\qquad \sqrt b=(3x^2+y)\sqrt y .$$

Hence

$$a-\sqrt b =(x^3+3xy)-(3x^2+y)\sqrt y =(x-\sqrt y)^3 .$$

The real cube-root function is injective on $\mathbb{R}$, so taking cube roots yields

$$\sqrt[3]{\,a-\sqrt b\,}=x-\sqrt y .$$

This result also holds for 'nth surd'.

If

$$\sqrt[n]{\,a+\sqrt b\,}=x+\sqrt y,$$

, then

$$(x-\sqrt y)^n=a-\sqrt b.$$

Example 1:

Solve for x,y
:

$$\sqrt[3]{37-30\sqrt{3}}=x-\sqrt{y}$$

Solution:

$$\sqrt[3]{37 - 30\sqrt{3}} = x - \sqrt{y}$$ $$\sqrt[3]{37 + 30\sqrt{3}} = x + \sqrt{y}$$ $$\sqrt[3]{(37{)}^2-{30}^2\cdot 3}=x^2-y$$

$$\sqrt[3]{1369 - 2700} = x^2 - y$$ $$\sqrt[3]{-1331} = x^2 - y \Rightarrow x^2 - y = -11$$ $$\boxed{{\displaystyle y=x^2+11}}$$

$$\sqrt[3]{37 - 30\sqrt{3}} = x - \sqrt{y}$$ $$37 - 30\sqrt{3} = x^3 - 3x^2\sqrt{y} + 3x y - y\sqrt{y}$$ $$x^3+3x(x^2+11)=37$$

$$x^3 + 3x^3 + 33x - 37 = 0$$ $$4x^3+33x-37=0$$

It has one root as x = 1, others imaginary.

$$\sqrt[3]{37-30\sqrt{3}}=x-\sqrt{y}=1-\sqrt{12}=1-2\sqrt{3}$$

This gives the values of x,y.

Example 2:

Let

$$\sqrt[3]{72-32\sqrt5}=x-\sqrt y .$$

Do it the same way as the example.

Let

$$\sqrt[3]{72-32\sqrt5}=x-\sqrt y \quad\Rightarrow\quad \sqrt[3]{72+32\sqrt5}=x+\sqrt y .$$

Multiply the two:

$$(x-\sqrt y)(x+\sqrt y)=x^2-y =\sqrt[3]{(72-32\sqrt5)(72+32\sqrt5)} =\sqrt[3]{72^2-(32\sqrt5)^2}=\sqrt[3]{64}=4,$$

so

$$y=x^2-4. \tag{1}$$

Now cube $x-\sqrt y$ and match rational/irrational parts:

$$(x-\sqrt y)^3=x^3+3xy-(3x^2+y)\sqrt y =72-32\sqrt5.$$

Hence

$$x^3+3xy=72,\qquad (3x^2+y)\sqrt y=32\sqrt5.$$

Using (1) in the first equation:

$$x^3+3x(x^2-4)=72 \;\Rightarrow\; 4x^3-12x=72 \;\Rightarrow\; x^3-3x-18=0.$$

Try integer roots: $x=3$ satisfies it. Then from (1),

$$y=x^2-4=9-4=5.$$

Finally,

$$\sqrt[3]{72-32\sqrt5}=x-\sqrt y=3-\sqrt5.$$

So $x=3,\; y=5$.

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