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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $a = \sqrt{3}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

One of the values of $\sqrt{24-70 i}+\sqrt{-24+70 i}$ is

$\sinh^{-1}(2) + \cosh^{-1}(2) - \tanh^{-1}\left(\frac{2}{3}\right) + \coth^{-1}(-2) = $

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

The square roots of the complex number $(-5-12i)$ are

$\sinh^{-1}\left(2^{3/2}\right)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module