$\text { The value of the integral } \int_\limits{\dfrac{1}{3}}^1 \dfrac{\left(x-x^3\right)^{\dfrac{1}{3}}}{x^4} d x \text { is }$

Let $I = \int_{\dfrac{1}{3}}^{1} \dfrac{(x-x^3)^{\dfrac{1}{3}}}{x^4} dx$.

Factor out $x^3$ from the term inside the cube root:

$I = \int_{\dfrac{1}{3}}^{1} \dfrac{(x^3(x^{-2}-1))^{\dfrac{1}{3}}}{x^4} dx = \int_{\dfrac{1}{3}}^{1} \dfrac{x(x^{-2}-1)^{\dfrac{1}{3}}}{x^4} dx = \int_{\dfrac{1}{3}}^{1} \dfrac{(x^{-2}-1)^{\dfrac{1}{3}}}{x^3} dx$.

Let $u = x^{-2} - 1$. Then $du = -2x^{-3} dx$, which implies $x^{-3} dx = -\dfrac{1}{2} du$.

Change the limits of integration:

When $x = \dfrac{1}{3}$, $u = (\dfrac{1}{3})^{-2} - 1 = 9 - 1 = 8$.

When $x = 1$, $u = 1^{-2} - 1 = 0$.

Substituting these into the integral:

$I = \int_{8}^{0} u^{\dfrac{1}{3}} (-\dfrac{1}{2}) du = \dfrac{1}{2} \int_{0}^{8} u^{\dfrac{1}{3}} du$.

Evaluating the integral:

$I = \dfrac{1}{2} [\dfrac{u^{\dfrac{4}{3}}}{\dfrac{4}{3}}]_{0}^{8} = \dfrac{1}{2} \times \dfrac{3}{4} [u^{\dfrac{4}{3}}]_{0}^{8} = \dfrac{3}{8} (8^{\dfrac{4}{3}} - 0) = \dfrac{3}{8} \times (2^3)^{\dfrac{4}{3}} = \dfrac{3}{8} \times 2^4 = \dfrac{3}{8} \times 16 = 3 \times 2 = 6$.

Answer: 6