Question 113

If x = 27 and $$\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{729}$$, the y =

as x = 27

$$\sqrt[3]{x}$$ = 3

we need to find $$\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{729}$$

3 + $$\sqrt[3]{y}$$ = 9

y = 216

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