The product of the roots of the equation $\sqrt[3]{8+x}+\sqrt[3]{8-x}=1$, is ______.

We have,

$\sqrt[3]{8+x}+\sqrt[3]{8-x}=1$

$(\sqrt[3]{8+x}+\sqrt[3]{8-x})^3=(1)^3$ [On cubing both sides]

$⇒8+x+8-x+3(64-x^2)^{1/3}(\frac{\sqrt[3]{8+x}+\sqrt[3]{8-x}}{1})=1$ 

$⇒16+3(64-x^2)^{1/3}=1$

$⇒15=-3(64-x^2)^{1/3}$

$⇒(64-x^2)^{1/3}=-5$

$⇒64-x^2=-125 ⇒x^2=189 ⇒ x=± 3\sqrt{21}$

Product of roots = $3\sqrt{21} × - 3\sqrt{21}=-189$