Let $y=2x-1.$ Then the given equation is equivalent to $\sqrt[4]{27+y}+\sqrt[4]{27-y}=3\sqrt[4]{2},\ (1)$ with $-27\le y\le 27.$ Raising $(1)$ to the fourth power and simplifying we get $6\sqrt{27^2-y^2}+4\sqrt[4]{27^2-y^2}(\sqrt{27+y}+\sqrt{27-y})=108\ (2).$ Squaring $(1)$ we get $\sqrt{27+y}+\sqrt{27-y}=9\sqrt{2}-2\sqrt[4]{27^2-y^2}.\ (3)$ Substituting $(3)$ into $(2)$ we get the quadratic $2a^2-36\sqrt{2}a+108=0,\ (4)$ where $a=\sqrt[4]{27^2-y^2}.\ (4)$ Solving the quadratic we find $a=9\sqrt{2}\pm6\sqrt{3},$ but because of $(4)$ only the negative sign leads to an acceptable solution. Therefore $a=9\sqrt{2}-6\sqrt{3}.$ Thus, $y=\pm\sqrt{58320\sqrt{6}-142155}$ and $x=\frac{1}{2}(y+1)=\frac{1}{2}\left(1\pm 27\sqrt{80\sqrt{6}-195}\right).$ The two solutions are approximately $13.72159$ and $-12.72159.$