Find the value of $x$

Let $u=(6x+28)^{1/3}$ and $v=(6x-28)^{1/3}$ . Then by the definition of $u$ and $v$ , and by the given equation, we obtain the following equations $u^3-v^3=56\quad\text{ and}\quad u-v=2.$ Let us solve this system of equations. Solving the second of these two equations for $u$ we obtain that $u=v+2$ . Substituting into the first equation we get $(v+2)^3-v^3=56.$ The latter equation simplifies to $6v^2+12v+8=56.$ Subtracting 56 and dividing 6, we get the equation $v^2+2v-8=0,$ with the solutions $v=-4$ and $v=2.$ Replacing $v$ in terms of $x,$ the following equations for $x$ can be obtained $(6x-28)^{1/3}=-4,$ or $(6x-28)^{1/3}=2.$ Solving these two equations for $x,$ we get that $x=6$ or $x=-6.$

$(6x+28)^{1/3}-(6x-28)^{1/3}=2\Rightarrow (6x+28)^{1/3}=2+(6x-28)^{1/3}\Rightarrow 6x+28=(2+(6x+28)^{1/3})^3$ $\Rightarrow 6x+28=8+12(6x-28)^{1/3}+6(6x-28)^{2/3}+((6x-28)^{1/3})^3\Rightarrow 48=6(6x-28)^{2/3}+12(6x-28)^{1/3}$ $\Rightarrow (6x-28)^{2/3}+2(6x-28)^{1/3}-8=0\Rightarrow ((6x-28)^{1/3}+4)((6x-28)^{1/3}-2)=0$ $\Rightarrow (6x-28)^{1/3}=2,-4\Rightarrow 6x-28=8,-64\Rightarrow 6x=36,-36\Rightarrow x=\pm 6$