$\int\limits_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x=$

6

$I = \int\limits_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x = \int\limits_{\frac{1}{3}}^1 \frac{x\left(x^{-2}-1\right)^{\frac{1}{3}}}{x^4} d x$

$\Rightarrow I = \int\limits_{\frac{1}{3}}^1 x^{-3} \left(x^{-2}-1\right)^{\frac{1}{3}} d x $

$y = x^{-2} - 1$

$dy = -2x^{-3} dx$

$\frac{dy}{-2} = x^{-3} dx$

$x → \frac{1}{3},  y → 8$

$x → 1 , y → 0$

$\Rightarrow I = \frac{-1}{2} \int\limits_8^0 y^{\frac{1}{3}} dy$

$= \frac{-1}{2} \times \frac{3}{4} [y^{\frac{4}{3}}]_8^0 = \frac{1}{2} \times \frac{3}{4} \times 16 = 6$

Option: 3