Answer: \(x=\frac8{27}\) or \(x=-\frac18\).

Rewrite the equation as a standard quadratic in the chosen variable, then solve and reject any value that does not satisfy the original form.

Let

\(m=x^{\frac13}.\)

Then \(x^{-\frac13}=\frac1m\), so the equation becomes

\(6m-\frac2m-1=0.\)

Multiplying by \(m\),

\(6m^2-m-2=0.\)

Factorising,

\((3m-2)(2m+1)=0.\)

Thus

\(m=\frac23\quad\text{or}\quad m=-\frac12.\)

Since \(m=x^{\frac13}\), cube both values:

\(x=\left(\frac23\right)^3=\frac8{27},\)

or

\(x=\left(-\frac12\right)^3=-\frac18.\)

Therefore the result matches the required answer.