Alg2.1.30

Find the cube root of $37-30\sqrt{3}\,$

Let

$\sqrt[3]{37-30\sqrt{3}}=x-\sqrt{y}\,$

Cubing on both sides, we get

$37-30\sqrt{3}=(x-\sqrt{y})^{3}=x^3-3x^2y+3xy-y\sqrt{y}=(x^3+3xy)-\sqrt{y}(3x^2+y)\,$

Comparing the values on bothsides

Let this equation be one

$x^3+3xy=37\,$

Let the Conjugate of the given value is

$x+\sqrt{y}=\sqrt[3]{37+30\sqrt{3}}\,$

Now

$(x+\sqrt{y})(x-\sqrt{y})=(\sqrt[3]{37+30\sqrt{3}})(\sqrt[3]{37-30\sqrt{3}})\,$

$x^2-y=\sqrt[3]{1369-2700}=\sqrt[3]{-1331}=-11\,$

Let this equation be 2

$x^2=y-11\,$

Substituting this value in equation one

$x^3+3x(x^2+11)=37\,$

$4x^3+33x-37=0\,$

x=1 satisfies this equation

By substituting this value in 2

$y=x^2+11=12\,$

Hence

$\sqrt[3]{37-30\sqrt{3}}=1+2\sqrt{3}\,$