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For $$k \neq 0$$; if $$a = \sqrt[3]{k} - \frac{1}{\sqrt[3]{k}}$$ then, $$a^3 + 3a =$$
$$\frac{k - 1}{k^2 + 1}$$
$$\frac{k^2 - 1}{k}$$
$$\frac{k - 1}{k^2}$$
$$\frac{k}{k^2 - 1}$$
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