Basic of Logarithms
hard

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

A

$2\sqrt 3 + \sqrt 2 $

B

$\sqrt 3 + 2\sqrt 2 $

C

$3\sqrt 3 + \sqrt 2 $

D

$\sqrt 3 + \sqrt 2 $

Solution

(d) Let $x = {(9\sqrt 3 + 11\sqrt 2 )^{1/3}}$

$ \Rightarrow $${x^3} = 9\sqrt 3 + 11\sqrt 2 $

$ = 6\sqrt 3 + 3\sqrt 3 + 9\sqrt 2 + 2\sqrt 2 $

$ = 3\sqrt 3 + 2\sqrt 2 + 6\sqrt 3 + 9\sqrt 2 $

$ = 3\sqrt 3 + 2\sqrt 2 + 3(2\sqrt 3 + 3\sqrt 2 )$

$ = 3\sqrt 3 + 2\sqrt 2 + 3\sqrt 2 \,.\,\sqrt 3 (\sqrt 2 + \sqrt 3 )$

$ = {(\sqrt 3 )^3} + {(\sqrt 2 )^3} + 3.\sqrt 2 .\sqrt 2 \,(\sqrt 3 + \sqrt 2 ) = {(\sqrt 3 + \sqrt 2 )^3}$

So, ${x^3} = {(\sqrt 3 + \sqrt 2 )^3}$

$x = \sqrt 3 + \sqrt 2 $.

Standard 11
Mathematics

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