Answer: \(x=-27\) or \(x=8\).

Let

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

Then the equation

\(x^{\frac13}+1=\dfrac{6}{x^{\frac13}}\)

becomes

\(u+1=\dfrac{6}{u}.\)

Since \(u=x^{1/3}\), we must have \(u\ne0\). Multiply by \(u\):

\(u^2+u=6.\)

So

\(u^2+u-6=0.\)

Factorise:

\((u+3)(u-2)=0.\)

Thus

\(u=-3\quad\text{or}\quad u=2.\)

Now return to \(x\). Since \(u=x^{1/3}\),

\(x=(-3)^3=-27\quad\text{or}\quad x=2^3=8.\)

Therefore

\(x=-27\quad\text{or}\quad x=8.\)