<s>[INST] Is it possible to prove equality between complicated algebraic expression and integer without mathematical programs or websites? 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?
 [/INST] 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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