Multiply x^{-3} with both numerator and denominator, then

8=\dfrac{1}{\dfrac{1}{x^{-3}} } = \dfrac{1\cdotp x^{-3} }{ \dfrac{1}{x^{-3}}\cdotp x^{-3}} = x^{-3}

Multiply x^3 with both sides of the equation

8x^3 = x^{-3}\cdotp x^3 = x^0=1

Divide both sides with 8

x^3 = \dfrac{1}{8}

Then,

x = \dfrac{1}{2} =0.5