Alg2.1.31

Find the cube root of $9\sqrt{3}+11\sqrt{2}\,$

Let

$\sqrt[3]{11\sqrt{2}+9\sqrt{3}}=\sqrt{x}+\sqrt{y}\,$

Cubing on both sides,

$11\sqrt{2}+9\sqrt{3}=(\sqrt{x}+\sqrt{y})^3=\sqrt{x}(x+3y)+\sqrt{y}(3x+y)\,$

Comparing the values both sides

Let this equation be 1

$\sqrt{x}(x+3y)=11\sqrt{2}\,$

Let this equation be 2

$\sqrt{y}(3x+y)=9\sqrt{3}\,$

Squaring 1 and 2

$x^3+9xy^2+6x^2y=242,9x^2y+y^3+6xy^2=243\,$

Subtracting first one from the second equations just above,

we get

$x-y=-1\,$

$x=y-1\,$

Substituting this value in equation 2,

$\sqrt{y}(3y-3+y)=9\sqrt{3}\,$

$(4y-3)\sqrt{y}=9\sqrt{3}\,$

Comparing bothsides,y=3 satisfies the equation,and hence value of x=2

Hence $\sqrt[3]{11\sqrt{2}+9\sqrt{3}}=\sqrt{2}+\sqrt{3}\,$