Square root Third Root

Let the given number be denoted as $n^3$

Hence,

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

Taking $3\sqrt{3}$ common from the R.H.S of the above equation,

$n^3 = 3\sqrt{3} \times ( 3 + \dfrac{11}{3} \sqrt{\dfrac{2}{3}} )$

Implies,

$n = \sqrt{3} \times ( 3 + \dfrac{11}{3} \sqrt{\dfrac{2}{3}} )^\frac{1}{3}$

Consider the number ( $3 + \dfrac{11}{3} \sqrt{\dfrac{2}{3}} )^\frac{1}{3}$ as $x + \sqrt{y}$

Therefore,

$x + \sqrt{y} = (3 + \dfrac{11}{3} \sqrt{\dfrac{2}{3}} )^\frac{1}{3}$ ......... (1)

$x - \sqrt{y} = (3 - \dfrac{11}{3} \sqrt{\dfrac{2}{3}} )^\frac{1}{3}$ ...........(2)

Multiplying (1) and (2),

$x^2 - y = ( 9 - \dfrac{121}{9} \times \dfrac{2}{3} )^\frac{1}{3}$

$x^2 - y = (\dfrac{1}{27})^\frac{1}{3}$

$x^2 - y = \dfrac{1}{3}$ .......... (3)

Cubing equation (1) (which I have not shown here) and comparing the rational parts of the L.H.S and R.H.S,

$x^3 + 3xy = 3$ .......... (4)

Equating (3) and (4) (which results in a cubic equation in $x$ ), we get,

$x = 1$ and $y = \dfrac{2}{3}$

Now back to the initial steps,

$n = \sqrt{3} \times (x + \sqrt{y})$

$n = \sqrt{3} \times (1 + \sqrt{\dfrac{2}{3}})$

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

$a =1 , b = 3 , c = 1 , d = 2$

Required answer is $1^1 + 3^2 + 1^3 + 2^4 = \boxed{27}$

$\large ~~~~\color{#3D99F6}{9}\sqrt3+\color{#D61F06}{11}\sqrt2\\\large=\color{#3D99F6}{3}\sqrt3~~~~+~~~~~~~~~~~\color{#D61F06}{9}\sqrt2~~~~~~~~~~~~~~+~~~~~~~~~~~\color{#3D99F6}{6}\sqrt3~~~~~~~~~~~ +~~~\color{#D61F06}{2}\sqrt2\\\large= (\sqrt3)^3~+~~~~~3*(\sqrt3)^2*\sqrt2~~~+~~3*\sqrt3*(\sqrt2)^2~~~+~(\sqrt2)^3\\\large =(\sqrt3+\sqrt2)^3=(a\sqrt b +c\sqrt d)^3 \\ \large a^1 + b^2 + c^3 + d^4=1^1 + 3^2 + 1^3 + 2^4 = \boxed{\Huge \color{#69047E}{27}}$