Question

Find the factor which will rationalise:
$\sqrt[3]{33}-\sqrt{2}$

Answer 1

We need to find the factor which when multiplied by $(\sqrt[3]{33}-\sqrt{2})$ will lead to a rational number

Let $x=3^{\frac{1}{3}}$ and $y=2^{\frac{1}{2}}$, then $x^6$ and $y^6$ are both rational

Also, $x^6-y^6=(x-y)(x^5+x^{4}y+x^3{y}^2+x^2{y}^3+x{y}^4+y^5)$

Hence, $x^5+x^{4}y+x^3{y}^2+x^2{y}^3+x{y}^4+y^5$ is the required factor.

Substituting the value for $x$ and $y$, then the required factor is

$3^{\frac{5}{3}}+3^{\frac{4}{3}}\cdot 2^{\frac{1}{2}}+3^{\frac{3}{3}}\cdot 2^{\frac{2}{2}}+3^{\frac{2}{3}}\cdot 2^{\frac{3}{2}}+3^{\frac{1}{3}}\cdot 2^{\frac{4}{2}}+2^{\frac{5}{2}}$

$\Rightarrow 3^{\frac{5}{3}}+3^{\frac{4}{3}}\cdot 2^{\frac{1}{2}}+6+3^{\frac{2}{3}}\cdot 2^{\frac{3}{2}}+4\cdot 3^{\frac{1}{3}}+2^{\frac{5}{2}}$