Question

If $\frac{\sqrt{2}+\sqrt{3}}{\sqrt[3]{2}-\sqrt[2]{3}}=a-b\sqrt{6}$ , find the values of a and b.

Answer 1

We have :
$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ = a $-$ b$\sqrt{6}$

Now, rationalising the denominator of the given number by multiplying
both the numerator and denominator with $3\sqrt{2}+2\sqrt{3}$, we get:

$$\begin{gathered} \frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} \\ =\frac{\sqrt{2}\left(3\sqrt{2}+2\sqrt{3}\right)+\sqrt{3}\left(3\sqrt{2}+2\sqrt{3}\right)}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ =\frac{6+2\sqrt{6}+3\sqrt{6}+6}{18-12} \\ =\frac{12+5\sqrt{6}}{6} \\ =\frac{12}{6}+\frac{5}{6}\sqrt{6} \\ =2+\frac{5}{6}\sqrt{6} \end{gathered}$$


∴ $2+\frac{5}{6}\sqrt{6}$ = a $-$ b$\sqrt{6}$
So, on comparing the LHS and RHS, we get:

a = 2 and b = $-\frac{5}{6}$