Question

Express with rational denominator:
$\frac{\sqrt{2}.\sqrt[3]{33}}{\sqrt[3]{33}+\sqrt{2}}$

Answer 1

We Know that

$(a+b)(a-b)=a^2-b^2$ .....(1)

$a^3-b^3=(a-b)(a^2+ab+b^2)$ .....(2)

Here in the problem the denominator has one power $\frac{1}{2}$ and one power $\frac{1}{3}$. First we rationalize power $\frac{1}{2}$ term ie.$\sqrt{2}$ and then rationalize power $\frac{1}{3}$ term.

For doing so we multiply both numerator and denominator with $\sqrt[3]{33}-\sqrt{2}$

$=\frac{\left(\sqrt{2}\sqrt[3]{33}\right)(\sqrt[3]{33}-\sqrt{2})}{(\sqrt[3]{33}{)}^2-2}$

$=\frac{(\sqrt[3]{33}{)}^2\sqrt{2}-2\sqrt[3]{33}}{(\sqrt[3]{33}{)}^2-2}$

Let $\sqrt[3]{33}=a,\sqrt{2}=b$

$=\frac{a^{2}b-b^{2}a}{a^2-b^2}$

Now multiplying both numerator and denominator with $(a^4+a^2{b}^2+b^4)$

$=\frac{(a^{2}b-b^{2}a)(a^4+a^2{b}^2+b^4)}{(a^2-b^2)(a^4+a^2{b}^2+b^4)}$

$=\frac{\left(ab\right)(a^5+a^3{b}^2+a{b}^4-a^{4}b-a^2{b}^3-b^5)}{a^6-b^6}$

Substituting $a,b$ in the above, we get

$=\frac{\left(\sqrt[3]{33}\sqrt{2}\right)\left[3\right(\sqrt[3]{33}{)}^2+6+4\sqrt[3]{33}-3\sqrt[3]{33}\sqrt{2}-2(\sqrt[3]{33}{)}^2\sqrt{2}-4\sqrt{2}]}{1}$

$=3^2{.2}^{\frac{1}{2}}-3^{\frac{5}{3}}.2+3^{\frac{4}{3}}{.2}^{\frac{3}{2}}-{3.2}^2+3^{\frac{2}{3}}{.2}^{\frac{5}{2}}-3^{\frac{1}{3}}{.2}^3$