84. Equation containing radicals. Solve the equation

$$
(6 x+28)^{\frac{1}{3}}-(6 x-28)^{\frac{1}{3}}=2
$$
84. If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c^{\star}$; therefore

$$
\begin{aligned}
(6 x+28)-(6 x-28)-8 & =3[(6 x+28)(6 x-28)(8)]^{\frac{1}{3}} \\
48 & =6\left(36 x^{2}-784\right)^{\frac{1}{3}} \\
512 & =36 x^{2}-784 \\
x^{2} & =36 \\
x & = \pm 6
\end{aligned}
$$

[F. E. Németh, S. S. M., 41, 291 (March 1941).]