Tolong dibantu jawab pake cara kk no 13

[tex] \sqrt[3]{ \frac{1}{ 8^{2x} } } [/tex] > [tex] \frac{ 64^{3x} }{ 2^{18x-36} } [/tex]
{ \frac{1}{ 8^{2x} } > ( [tex] \frac{ 64^{3x} }{ 2^{18x-36} } [/tex] ) [tex]^{3} [/tex]
1/2^3(2x) > 64^9x/2^54x-108
2^-6x > 2^54x-(54x-108)
-6x > 54x-(54x-108)
-6x > 54x-54x+108
-6x > 108
x < 108/-6
x < -18  (E)