#### If $\sqrt{2}= 1\cdot 414$ and $\sqrt{3}= 1\cdot 732$ then find the value of $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$

Solution.   Given that :
$\sqrt{2}= 1\cdot 414$,$\sqrt{3}= 1\cdot 732$
$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$
$= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}+\frac{3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}+2\sqrt{2} \right )\left (3\sqrt{3}-2\sqrt{2} \right )}$
$= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )+3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}$
Using   (a – b) (a + b) = a2 – b2
$= \frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}$
$= \frac{21\sqrt{3}+2\sqrt{2}}{27-8}$
$= \frac{21\sqrt{3}+2\sqrt{2}}{19}$

Putting the given values,$= \frac{21\left ( 1\cdot 732 \right )+2\left ( 1\cdot 414 \right )}{19}$
$=\frac{39\cdot 2014}{19}$
= 2.0632