You can use the identity $ x - a = \left( \sqrt[3]{x} - \sqrt[3]{a} \right) \left( \sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} \right), $ which is derived from the following identity: $ a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}). $ Then $ \forall x \in \mathbb{R} \setminus \{ a \}: \quad \sqrt[3]{x} - \sqrt[3]{a} = \frac{x - a}{\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}}}. $ Now, fix $ \epsilon > 0 $. Choose $ x \in \mathbb{R} $ so that $ |x - a| < \min \left( \frac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right). $ As $ a > 0 $, having $ |x - a| < \dfrac{a}{2} $ ensures that $ x > 0 $, and so $ \sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} > \sqrt[3]{a^{2}}. $ Next, having $ |x - a| < \sqrt[3]{a^{2}} \cdot \epsilon $ yields $ \left| \sqrt[3]{x} - \sqrt[3]{a} \right| < \epsilon. $ You can therefore set $ \delta := \min \left( \dfrac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right) $.