If 2x + \(\frac{1}{3x}\) + 1 = 0

find \(\sqrt[3]{{27x}^{3}\;+\;\frac{1}{{8x}^{3}}}\).

\(\frac{3}{2}\)

2x + \(\frac{1}{3x}\) + 1 = 0

2x + \(\frac{1}{3x}\) = -1

Multiply by \(\frac{3}{2}\) in equation, we get

⇒ 3x + \(\frac{1}{2x}\) = \(\frac{-3}{2}\)

Cubing both side,

⇒ 27x³ + \(\frac{1}{{8x}^{3}}\) = \(\frac{-27}{8}\) - 3 \(\times\) \(\frac{3}{2}\) (\(\frac{-3}{2}\))

                       =  \(\frac{-27}{8}\) + \(\frac{27}{4}\)

                       =   \(\frac{-27+54}{8}\) = \(\frac{27}{8}\)

Now,

\(\sqrt[3]{{27x}^{3}+\frac{1}{{8x}^{3}}}\) = \(\sqrt[3]\frac{27}{8}\) = \(\frac{3}{2}\)