The cube root of $9\sqrt 3 + 11\sqrt 2 $ is
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 $
Similar Questions
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $
${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $
The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$ is
The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$ is
${a^{m{{\log }_a}n}} = $
${a^{m{{\log }_a}n}} = $
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is