Vieta?

use the identity $a^3+b^3 = (a+b)^3 - 3ab(a+b)$ twice, i.e $x+\dfrac{1}{x}= \left(\sqrt[3]{x}+\dfrac{1}{\sqrt[3]{x}} \right)^3-3\sqrt[3]{x}\dfrac{1}{\sqrt[3]{x}}\left(\sqrt[3]{x}+\dfrac{1}{\sqrt[3]{x}} \right)=3^3-3\times 3=18\\ x^3+\dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right)^3-3 x \dfrac{1}{x} \left(x+\dfrac{1}{x}\right) = 18^3-3\times 18 =\boxed{5778}$

$\begin{aligned} \left(\sqrt[3]x+\frac 1{\sqrt[3]x}\right)^3 & = 3^3 \\ x + 3\sqrt[3]x+\frac 3{\sqrt[3]x} + \frac 1x & = 27 \\ \implies x + \frac 1x & = 27 - 3\left(\sqrt[3]x+\frac 1{\sqrt[3]x}\right) \\ & = 27 - 9 = 18 \end{aligned}$

Similarly,

$\begin{aligned} \left(x+\frac 1x\right)^3 & = 18^3 \\ x^3 + 3\left(x+\frac 1x\right) + \frac 1{x^3} & = 5832 \\ \implies x^3 + \frac 1{x^3} & = 5832 - 3\left(x+\frac 1x\right) \\ & = 5832 - 3(18) = \boxed{5778} \end{aligned}$

We have $(x+\frac{1}{x})^3 = x^3 + 3 \cdot x^2 \cdot \frac{1}{x} + 3 \cdot x \cdot \frac{1}{x^2} + \frac{1}{x^3} = x^3 + \frac{1}{x^3} + 3 (x+\frac{1}{x}) \implies x^3 + \frac{1}{x^3} = (x+\frac{1}{x})^3 - 3(x+\frac{1}{x})$ .

So $x + \frac{1}{x} = (\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}})^3 - 3(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}) = 3^3 - 3 \cdot 3 = 18$

So $x^3 + \frac{1}{x^3} = (x+\frac{1}{x})^3 - 3(x+\frac{1}{x}) = 18^3 - 3 \cdot 18 = \boxed{5778}$

$\text{FullSimplify}\left[x^3+\frac{1}{x^3}\text{/.}\, \text{Solve}\left[\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}=3\right]\right] \Rightarrow \{5778,5778\}$

$x = 9\pm4 \sqrt{5}$

Substitute $y=x^{\frac13}$ giving $y^2-3 y+1=0$ with the solution $y=\frac{1}{2} \left(3\pm\sqrt{5}\right)$ , which substitutes easily into $x^3+\frac{1}{x^3}$ giving $y^9+\frac{1}{y^9}$ with a final value of 5778.

By the way, that is the surface temperature of the Sun in Kelvin.