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The ascending order of $$\sqrt[3]{5}, \sqrt[4]{5}$$ and $$\sqrt[6]{5}$$ is
$$\sqrt[3]{5} < \sqrt[4]{5} < \sqrt[6]{5}$$
$$\sqrt[4]{5} < \sqrt[3]{5} < \sqrt[6]{5}$$
$$\sqrt[6]{5} < \sqrt[4]{5} < \sqrt[3]{5}$$
$$\sqrt[6]{5} < \sqrt[3]{5} < \sqrt[4]{5}$$
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