Is the product of two radical expressions equal to six?

[anemone]

Here is this week's POTW:

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Prove that $\sqrt{4+\sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}} \cdot \sqrt{8+2\sqrt[3]{19+3\sqrt{33}}+2\sqrt[3]{19-3\sqrt{33}}}=6$.

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[bob012345]

Please clarify the problem. Wolfram Alpha says this is not 6. Thanks.

 

[I like Serena]

Please clarify the problem. Wolfram Alpha says this is not 6. Thanks.

I've checked with W|A and it confirms that it is really 6.
Note that W|A treats every input as complex valued and takes the principle complex root by default, while in this case the real valued root is intended. That is, we need that $\sqrt[3]{-1}=-1$ instead of $\sqrt[3]{-1}=e^{i\pi/3}$.

 

[anemone]

I've checked with W|A and it confirms that it is really 6.
Note that W|A treats every input as complex valued and takes the principle complex root by default, while in this case the real valued root is intended. That is, we need that $\sqrt[3]{-1}=-1$ instead of $\sqrt[3]{-1}=e^{i\pi/3}$.

Thanks @I like Serena for the reply!

 

[anuttarasammyak]

All the terms except
(-26-6\sqrt{33})^{1/3}
are real. I do not believe this equation holds. Should I interpret it with real coefficient -1
-(26+6\sqrt{33})^{1/3}
not with complex coefficients, i.e.
e^{\pi i/3},e^{-\pi i/3}

[EDIT] I got it in the post #4. Thanks.

 

[malawi_glenn]

All the terms except
(-26-6\sqrt{33})^{1/3}
are real. I do not believe this equation holds.

 

[anuttarasammyak]

This is just to know I got the question properly. LHS of the to be proved equation is
L:=2^{2/3}[2^{5/3}+(a-13)^{1/3}-(a+13)^{1/3}]^{1/2}[2^2+(19+a)^{1/3}+(19-a)^{1/3}]^{1/2}
where
a=3\sqrt{33}=17.23...
We may make use of the relations
(a+13)(a-13)=a^2-13^2=128=2^7
(a+19)(19-a)=19^2-a^2=64=2^6
For an example
[2^2+(19+a)^{1/3}+(19-a)^{1/3}]^{1/2}=(19+a)^{1/6}+(19-a)^{1/6}

 

[bob012345]

Is this problem workable without infinite series expansions?

 

[bob012345]

So far after spending too many hours on this I wonder if there is a doable closed form proof. Of course one can evaluate this easily with computer tools like Wolfram Alpha but proof is not the same thing. The cube roots are all irrational numbers.

I did notice that

$$\sqrt[3]{-26+6\sqrt{33}}*\sqrt[3]{-26-6\sqrt{33}} = -8$$
and $$\sqrt[3]{19+3\sqrt{33}}*\sqrt[3]{19-3\sqrt{33}}=4$$

 

[bob012345]

I have tried everything. When will the proof be posted?

 

[anemone]

I am so sorry for the late reply to this POTW's thread as I have been so extremely busy with work.

Here is the solution of other and I hope that readers will appreciate the beauty of the solution as much as I do.

Spoiler

Let $x=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}$ so that we get $x^3-3\sqrt[3]{a^2-b}x-2a=0$.

Now, let $p=\sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}$ and $q=\sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}$. We then obtained $q^3-12q-38=0$ and $p^3+24p+52=0$.

Then what we are required to prove becomes $(4+p)(4+q)=18$.

It is then not hard to prove that both $q^3-12q-38=0$ and $p^3+24p+52=0$ have only one real roots.

That means $4+q$ and $4+p$ are the only real root of the polynomials $f(x)=(x-4)^3-12(x-4)-38=x^3-12x^2+36x-54$ and $g(x)=(x-4)^3+24(x-4)+52=x^3-12x^2+72x-108$.

So we have $f\left(\dfrac{18}{x}\right)=-\dfrac{54}{x^3}g(x)$, i.e. $f\left(\dfrac{18}{4+p}\right)=g(4+q)=0$.

Hence, $\dfrac{18}{4+p}=4+q$ and the proof follows.

 

[anuttarasammyak]

For confirmation of post#11

 

[bob012345]

Thanks. Before I look at it I want to report on the direction I took but I could not finish it.

Spoiler

$$\sqrt{4+\sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}} \cdot \sqrt{8+2\sqrt[3]{19+3\sqrt{33}}+2\sqrt[3]{19-3\sqrt{33}}}=6$$

Let $$z= \sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}$$ and
$$z' = \sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}$$

we can write it as

$$\sqrt{4+z} \cdot \sqrt{8+2z'}=6$$

or $$32 + 8z + 8z' + 2zz' = 36$$

solving for z' we get

$$z' = \frac{2 - 4z}{4 + z}$$

we put in z in the expression

$$z' = \frac{2 - 4\sqrt[3]{-26+6\sqrt{33}} -4\sqrt[3]{-26-6\sqrt{33}})}{4 + \sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}}$$

then try and simplify. Wolfram Alpha says it does simplify to exactly z' but so far I have not been able to do it.

[\SPOILER]

 

[bob012345]

I finally have finished my proof in a direct form.

Spoiler

given

$$\sqrt{4+\sqrt[3]{-26+6\sqrt{33}}+\sqrt[3]{-26-6\sqrt{33}}} \cdot \sqrt{8+2\sqrt[3]{19+3\sqrt{33}}+2\sqrt[3]{19-3\sqrt{33}}}=6$$

write it as
$$\sqrt{4+ a + b} \sqrt{8+2c + 2d} = 6$$

expand
$$ \sqrt{32 + 8(a + b + c + d) + 2(ac + ad + bc + bd)} = 6$$

then

$a = \sqrt[3]{-26 + 6\sqrt{33}}\hspace{0.5cm} b= \sqrt[3]{-26 - 6\sqrt{33}}\hspace{0.5cm}c = \sqrt[3]{19 + 3\sqrt{33}} \hspace{0.5cm}d= \sqrt[3]{19 - 3\sqrt{33}}$

Details of the working out of these expressions are in the appendix below.

$ac = 1 + \sqrt{33}\hspace{0.5cm}bd = 1 - \sqrt{33}\hspace{0.5cm}ad = -4\sqrt[3]{17-3\sqrt{33}}\hspace{0.5cm}bc = -4\sqrt[3]{17+3\sqrt{33}}$

Evaluating the sum of terms it is convenient to add $a+d$ and $b+c$ separately.

$a + d = \sqrt[3]{17+3\sqrt{33}}\hspace{0.5cm}b + c = \sqrt[3]{17-3\sqrt{33}}$

therefore

$$ \sqrt{32 + 8\sqrt[3]{17+3\sqrt{33}}) +8\sqrt[3]{17-3\sqrt{33}}+ 2\left( 1 + \sqrt{33} + 1 - \sqrt{33} -4\sqrt[3]{17+3\sqrt{33}} -4\sqrt[3]{17-3\sqrt{33}}\right)} = 6$$

giving

$$ \sqrt{32 + 4} = \sqrt{36} = 6$$

Appendix:
Again;
$a = \sqrt[3]{-26 + 6\sqrt{33}}\hspace{0.5cm} b= \sqrt[3]{-26 - 6\sqrt{33}}\hspace{0.5cm}c = \sqrt[3]{19 + 3\sqrt{33}} \hspace{0.5cm}d= \sqrt[3]{19 - 3\sqrt{33}}$

$$ac = \sqrt[3]{-26 + 6\sqrt{33}} \sqrt[3]{19 + 3\sqrt{33}} = \sqrt[3]{-26*19 +18*33 +(6*19-3*26)\sqrt{33}} = \sqrt[3]{100 + 36\sqrt{33}} = \sqrt[3]{(1 + \sqrt{33})^3} = 1 + \sqrt{33}$$
likewise
$$bd = \sqrt[3]{-26 - 6\sqrt{33}} \sqrt[3]{19 - 3\sqrt{33}} = \sqrt[3]{-26*19 +18*33 +(-6*19+3*26)\sqrt{33}} = \sqrt[3]{100 - 36\sqrt{33}} = \sqrt[3]{(1 - \sqrt{33})^3} = 1 - \sqrt{33}$$

$$ad = \sqrt[3]{-26 + 6\sqrt{33}} \sqrt[3]{19 - 3\sqrt{33}} = \sqrt[3]{-26*19 -18*33 +(6*19+3*26)\sqrt{33}} = \sqrt[3]{-1088 + 192\sqrt{33}} = \sqrt[3]{64(-1 7+ 3\sqrt{33})} = -4\sqrt[3]{17 - 3\sqrt{33}}$$
likewise
$$bc = \sqrt[3]{-26 - 6\sqrt{33}} \sqrt[3]{19 + 3\sqrt{33}} = \sqrt[3]{-26*19 -18*33 -(6*19+3*26)\sqrt{33}} = \sqrt[3]{-1088 - 192\sqrt{33}} = \sqrt[3]{64(-1 7- 3\sqrt{33})} = -4\sqrt[3]{17 + 3\sqrt{33}}$$

Now for the sum $(a + d)$ we use the relation $(a + d)^3 = a^3 + d^3 +3a^2d + 3ad^2$

$a^2d = a (ad) = -4\sqrt[3]{(-26 + 6\sqrt{33})(17 - 3\sqrt{33})} = -4\sqrt[3]{-1036 + 180\sqrt{33}} = \sqrt[3]{66304 - 11520\sqrt{33}} = \sqrt[3]{(28 - 4\sqrt{33})^3} = 28 - 4\sqrt{33}$

$ad^2 = (ad)d = -4\sqrt[3]{(17 - 3\sqrt{33})(19 - 3\sqrt{33})} = -4\sqrt[3]{620 - 108\sqrt{33}} = -\sqrt[3]{39680 - 6912\sqrt{33}} = \sqrt[3]{(-20 + 4\sqrt{33})^3} = -20 + 4\sqrt{33}$

so $(a + d)^3 = -26 + 6\sqrt{33} + 19 - 3\sqrt{33} +3(28 - 4\sqrt{33}) + 3( -20 + 4\sqrt{33}) = -7 +3\sqrt{33} +84 -12\sqrt{33} -60 +12\sqrt{33} = 17 +3\sqrt{33}$

thus $(a + d) = \sqrt[3]{ 17 +3\sqrt{33}}$

Likewise for the sum $(b + c)$ we use the relation $(b + c)^3 = b^3 + c^3 +3b^2c + 3bc^2$

$b^2c = b (bc) = -4\sqrt[3]{(-26 - 6\sqrt{33})(17 + 3\sqrt{33})} = -4\sqrt[3]{-1036 - 180\sqrt{33}} = \sqrt[3]{66304 + 11520\sqrt{33}} = \sqrt[3]{(28 + 4\sqrt{33})^3} = 28 + 4\sqrt{33}$

$bc^2 = (bc)c = -4\sqrt[3]{(17 + 3\sqrt{33})(19 + 3\sqrt{33})} = -4\sqrt[3]{620 + 108\sqrt{33}} = -\sqrt[3]{39680 + 6912\sqrt{33}} = \sqrt[3]{(-20 - 4\sqrt{33})^3} = -20 - 4\sqrt{33}$

so $(b + c)^3 = -26 - 6\sqrt{33} + 19 + 3\sqrt{33} +3(28 + 4\sqrt{33}) + 3( -20 - 4\sqrt{33}) = -7 -3\sqrt{33} +84 +12\sqrt{33} -60 -12\sqrt{33} = 17 -3\sqrt{33}$

thus $(b + c) = \sqrt[3]{ 17 -3\sqrt{33}}$

[\SPOILER]