$f(x) = x^{\frac{1}{3} } \; \; \;\; \; \; \therefore f ' (x) = \frac{1}{3} x^{-2/3}$
$\implies f(x + \triangle x) = (x + \triangle x)^ {\frac{1}{3}}$
$x = 0.008$ and $\triangle x = 0.001$
$f(x+ \triangle x) = \triangle x . f'(x) + f(x)$
$\implies (x + \triangle x)^{1/3} = \triangle x . (\frac{1}{3} x ^{-2/3}) + x^{1/3}$
$(0.008 + 0.001)^{1/3} = \frac{0.001}{3(0.008)^{2/3}} + (0.008)^{1/3}$
$\begin{align*}\implies (0.009)^{1/3} & = \frac{0.001+0.024}{0.12} = \frac{0.025}{0.012} \\ & = 0.2083 \end{align*}$