Question

Express with rational denominator:
$\frac{\sqrt{8}+\sqrt[3]{34}}{\sqrt{8}-\sqrt[3]{34}}$

Answer 1

Let $\sqrt{8}=2^{\frac{3}{2}}=a$ ,$\sqrt[3]{34}=2^{\frac{2}{3}}=b$

Here in the problem the denominator has one power $\frac{3}{2}$ and one power $\frac{2}{3}$.First we rationalize power $\frac{3}{2}$ term ie.$\sqrt{8}$ and then rationalize power $\frac{2}{3}$ term.Basically we need to get $a^6-b^6$ form.

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

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

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

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

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

substituting a,b

$=\frac{2^9+2^8{.2}^{\frac{1}{3}}+2^6{.2}^{\frac{1}{2}}+2^9{.2}^{\frac{1}{6}}+2^6{.2}^{\frac{1}{2}}+2^5{.2}^{\frac{5}{6}}+2^4}{2^4.31}$

$=\frac{2^5+2^{\frac{31}{6}}+2^{\frac{26}{6}}+2^{\frac{21}{6}}+2^{\frac{16}{6}}+2^{\frac{11}{6}}+1}{31}$