Ok

$\displaystyle x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2} $

$\displaystyle y = 8 \cos ^3 \theta \Rightarrow \cos \theta = \frac{\sqrt[3]{y}}{2}$

$\displaystyle \sin ^2 \theta + \cos ^2 \theta = 1 = \frac{\sqrt[3]{x^2}}{4} + \frac{\sqrt[3]{y^2}}{4} $

$\displaystyle 0 \leq x \leq 8 \;\;\; and \;\;\; -8 < y \leq 8 $

$\displaystyle y =\sqrt{3} x - 8 \Rightarrow y+8 = \sqrt{3} x\Rightarrow (y+8)^2 = 3x^2 $

so now we have to curves see this

link which solve these two curves and give x = $\displaystyle 3\sqrt{3} $ which is equal to 5.196