The cube root of 9sqrt{3} + 11sqrt{2} is | vedclass

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(D) Let $x = (9\sqrt{3} + 11\sqrt{2})^{1/3}$.Then $x^3 = 9\sqrt{3} + 11\sqrt{2}$.We know that $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.Let $a = \sqrt{3}$

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$\sqrt{3} + \sqrt{2}$

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(D) Let $x = (9\sqrt{3} + 11\sqrt{2})^{1/3}$.Then $x^3 = 9\sqrt{3} + 11\sqrt{2}$.We know that $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.Let $a = \sqrt{3}$ and $b = \sqrt{2}$.Then $a^3 = (\sqrt{3})^3 = 3\sqrt{3}$ and $b^3 = (\sqrt{2})^3 = 2\sqrt{2}$.$a^3 + b^3 = 3\sqrt{3} + 2\sqrt{2}$.$3ab(a + b) = 3(\sqrt{3})(\sqrt{2})(\sqrt{3} + \sqrt{2}) = 3\sqrt{6}(\sqrt{3} + \sqrt{2}) = 3(3\sqrt{2} + 2\sqrt{3}) = 9\sqrt{2} + 6\sqrt{3}$.Adding these,$x^3 = (3\sqrt{3} + 2\sqrt{2}) + (9\sqrt{2} + 6\sqrt{3}) = 9\sqrt{3} + 11\sqrt{2}$.Thus,$x^3 = (\sqrt{3} + \sqrt{2})^3$.Therefore,$x = \sqrt{3} + \sqrt{2}$.

The cube root of $9\sqrt{3} + 11\sqrt{2}$ is

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