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Prove:

$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}+\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=3$

From WolframAlpha:

First:

by

https://www.wolframalpha.com/input/?i=cubrt%283%2B10isqrt%281%2F27%29%29

$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}=(1.499...9...)+(0.288675134...)i$

Second:

by

https://www.wolframalpha.com/input/?i=cubrt%283-10isqrt%281%2F27%29%29

$\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=(1.499...9...)-(0.288675134...)i$

By first and second:

$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}+\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=3$

So is there way to prove this without mathematical programs or websites?

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    $\begingroup$ My first try, which might not pay off, would be to expressing the LHS as $(a + bi)^{(1/3)} + (a - bi)^{(1/3)}$, and let these two terms be represented by $r$ and $s$ respectively. Then, I would try to prove that $r^3 + 3rs(r + s) + s^3 = (r + s)^3 = 3^3 = 27.$ $\endgroup$ Sep 4, 2021 at 15:46
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    $\begingroup$ @user2661923 I tried this method ...it produces similar algebraic expression after simplifications. $\endgroup$ Sep 4, 2021 at 15:48
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    $\begingroup$ Please edit your question to show your work. $\endgroup$ Sep 4, 2021 at 15:49
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    $\begingroup$ Just to tease you: there is a way to prove it using a calculation that is (just about) simple enough to do mentally, and very quick and easy to do on paper. Strictly speaking, that is an answer to your question. :) $\endgroup$ Sep 4, 2021 at 19:04

3 Answers 3

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Recall the algebraic identity $$a^3+b^3 + c^3 - 3abc = \frac12(a+b+c)((a-b)^2 + (b-c)^2+(c-a)^2)$$ Whenever $a + b + c = 0$, we have $a^3 + b^3 + c^3 = 3abc$.

Let $\rho = 3 \pm \frac{10}{\sqrt{27}}i$ and $u = \sqrt[3]{\rho} + \sqrt[3]{\bar{\rho}} = 2\Re(\sqrt[3]{\rho})$. Substitute $a,b,c$ by $\rho$, $\bar{\rho}$ and $-u$, we obtain

$$\rho_{+} + \rho_{-} - u^3 = 3\sqrt[3]{\rho_{+}\rho_{+}} u$$ Since $\rho_{+} + \rho_{-} = 6$ and $\rho_{+}\rho_{-} = 3^2 + \frac{100}{27} = \left(\frac{7}{3}\right)^3$, this reduces to $$6-u^3 = 7u \iff u^3-7u-6 = (u-3)(u+1)(u+2) = 0$$ Since $\frac{10}{\sqrt{27}} < 3$, $$\arg(\rho_{+}) \in (0,\frac{\pi}{4}) \implies \arg(\sqrt[3]{\rho_{+}}) \in (0,\frac{\pi}{12}) \implies \Re( \sqrt[3]{\rho_{+}}) > 0$$ This means $u > 0$ and hence $u$ can only be $3$.

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Hint:

What happens if you generate an equation that looks like

$x^3 - 7x - 6 = 0.$

In addition to being able to immediately conclude that $3$ is one of the roots, you can use the resultant quadratic equation to determine the other two roots. Since you know that the original evaluation has to be one of the $3$ roots, you can then try to eliminate the two other roots from consideration.

If this succeeds, you will have proven that the expression must evaluate to $3$.


In the remainder of this answer, it is presumed that cube root represents the principal cube root, so that
$\left(re^{i\theta}\right) ~: -\pi < \theta \leq \pi$ will have the cube root of
$r^{(1/3)}e^{i\theta/3}.$


Hint-2

When I looked to derive and eliminate the other two roots, the first thing that I noticed was that the other two roots were also real numbers. This simplified eliminating the other two roots.

Suppose that $a = 3 + i[10\sqrt{27}].$
Further suppose that $b = 3 - i[10\sqrt{27}].$

Then $|a| = |b| \approx 3.56.$

This implies that $|a^{(1/3)}| = |b^{(1/3)}| \approx (3.56)^{1/3} \approx 1.52.$

This means that the original expression (i.e. the cube root of two different terms) must consist of a pair of complex conjugates, each of whom has a modulus of about $1.52.$

Now, you can consider the approximate ratio of the real portion of $a^{(1/3)}$ versus the imaginary portion of $a^{(1/3)}.$ You can use this type of analysis to show that the other two (candidate) roots are untenable, because they each have an untenable modulus. That is, geometrically, the modulus of either of the other two (candidate) roots is untenable versus the modulus of $|a^{(1/3)} + b^{(1/3)}|.$

Hint-3

An alternative approach is to reason that (re Hint-2 above), $a^{(1/3)}$ is in the 1st quadrant, and $b^{(1/3)}$ is in the 4th quadrant. Therefore, if their sum is going to intersect the Real Axis, the intersection can not take place on the negative side of the Real Axis.

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$$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}} +\sqrt[3]{3-10i\sqrt{\frac{1}{27}}} \; \overset?= \; 3$$


Let $u = \sqrt[3]{3+10i\sqrt{\frac{1}{27}}}$

Let $v = \sqrt[3]{3-10i\sqrt{\frac{1}{27}}}$

Note

  • $uv = \sqrt[3]{9 + \frac{100}{27}} = \frac 73$
  • $u^3 + v^3 = $6

Let $x = u+v$

\begin{align} (u+v)^3 &= u^3 +3u^2v + 3uv^2 + v^3 \\ &= (u^3 + v^3) +3uv(u+v) \\ &= 6 + 7(u+v) \\ \hline x^3 &= 6+ 7x \\ x^3-7x-6 &= 0 \\ (x+1)(x+2)(x-3) &= 0 \\ \hline u + v &\in \{-1, -2, 3\} \end{align}

You need to remember that $u = \sqrt[3]{3+10i\sqrt{\frac{1}{27}}}$ and $v = \sqrt[3]{3-10i\sqrt{\frac{1}{27}}}$ each have three distinct complex-number values. So, depending on which complex-valued cube roots you choose

$$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}} +\sqrt[3]{3-10i\sqrt{\frac{1}{27}}} = 3, -1, \text{ or } -2$$


Done numerically,

$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}} = \left\{ \begin{array}{c} -1 + 1.1547 i \\ -0.5 - 1.44338 i \\ 1.5 + 0.288675 i \end{array} \right.$

$\sqrt[3]{3-10i\sqrt{\frac{1}{27}}} = \left\{ \begin{array}{c} -1 - 1.1547 i \\ -0.5 + 1.44338 i \\ 1.5 - 0.288675 i \end{array} \right.$

and you can see, depending on which roots you choose \begin{align} (-1 + 1.1547 i) + (-1 - 1.1547 i) &= -2 \\ (-0.5 + 1.44338 i) + (-0.5 - 1.44338 i) &= -1 \\ ( 1.5 + 0.288675 i) + (1.5 - 0.288675 i) &= 3 \end{align}

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