Question

Find the value of $\sqrt[3]{3212+\sqrt[3]{364}}$.

Answer 1

Here, starting from the innermost cube we have

$\sqrt[3]{364}=\sqrt[3]{34\times 4\times 4}=4$

$\sqrt[3]{3212+\sqrt[3]{364}}=\sqrt[3]{3212+4}=\sqrt[3]{3216}$

Now, $\sqrt[3]{3216}=\sqrt[3]{36\times 6\times 6}=6$

Hence, $\sqrt[3]{3212+\sqrt[3]{364}}=6.$