The approximate value of $(0.007)^{1 / 3}$ is

Let $f(x) = x^{1/3}$. We want to find the approximate value of $(0.007)^{1/3}$.

Consider $x = 0.008$ and $\Delta x = -0.001$. Then $x + \Delta x = 0.007$.

Using the linear approximation formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$, we have $f'(x) = \dfrac{1}{3} x^{-2/3} = \dfrac{1}{3 x^{2/3}}$.

For $x = 0.008$, $f(0.008) = (0.008)^{1/3} = 0.2$.

$f'(0.008) = \dfrac{1}{3 (0.008)^{2/3}} = \dfrac{1}{3 (0.2)^2} = \dfrac{1}{3 \times 0.04} = \dfrac{1}{0.12} = \dfrac{100}{12} = \dfrac{25}{3}$.

Substituting these values into the approximation formula:

$(0.007)^{1/3} \approx 0.2 + \left(\dfrac{25}{3}\right) (-0.001) = 0.2 - \dfrac{0.025}{3} = \dfrac{0.6 - 0.025}{3} = \dfrac{0.575}{3} = \dfrac{575}{3000}$.

Simplifying the fraction $\dfrac{575}{3000}$ by dividing by 25:

$\dfrac{575 \div 25}{3000 \div 25} = \dfrac{23}{120}$.

Answer: $\dfrac{23}{120}$