Question

Express with rational denominator:
$\frac{\sqrt[3]{33}}{\sqrt{3}+\sqrt[6]{69}}$

Answer 1

Here $\sqrt[6]{69}=9^{\frac{1}{6}}=3^{2.\frac{1}{6}}=3^{\frac{1}{3}}$

$=\frac{3^{\frac{1}{3}}}{3^{\frac{1}{2}}+3^{\frac{1}{3}}}$

We Know that $(a+b)(a-b)=a^2-b^2$ ,$a^3-b^3=(a-b)(a^2+ab+b^2)$

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 i.e. $\sqrt{3}$ and then rationalize power $\frac{1}{3}$ term. Basically we need to get $a^6-b^6$ form.

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

$=\frac{a}{a+b}$

Multiplying numerator and denominator with $(a-b)$

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

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

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

$=\frac{a^6+a^4{b}^2+a^2{b}^4-a^{5}b-a^3{b}^3-a{b}^5}{a^6-b^6}$

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

$=\frac{3^2+3^2{.3}^{\frac{1}{3}}+3^2{.3}^{\frac{2}{3}}-3^2{.3}^{\frac{1}{6}}-3^2{.3}^{\frac{1}{2}}-3^2{.3}^{\frac{5}{6}}}{-3^2.2}$

$=\frac{1}{2}(3^{\frac{5}{6}}-3^{\frac{2}{3}}+3^{\frac{1}{2}}-3^{\frac{1}{3}}+3^{\frac{1}{6}}-1)$