Prove That
$\displaystyle \frac{1}{\sqrt7-\sqrt6}=\frac{3}{\sqrt6-\sqrt3}+\frac{4}{\sqrt7+\sqrt3}$
we can prove this by showing we can make the LHS and RHS look the same.
take the left hand side and multiply it by $\displaystyle \frac {(\sqrt {6} - \sqrt {3})(\sqrt {7} + \sqrt {3})}{(\sqrt {6} - \sqrt {3})(\sqrt {7} + \sqrt {3})}$ (which is 1, so you're not changing anything). simplify the expression
separately, take the right hand side, combine the fractions, then multiply by $\displaystyle \frac {\sqrt {7} - \sqrt {6}}{\sqrt {7} - \sqrt {6}}$ (which is 1, so you're not changing anything). simplify the expression and note that it is the same expression you got before when manipulating the left hand side. thus the equation is proved
now, how did i know what to multiply by?
good guess...weird if it's a guess, but good
yes, you multiply each side by the LCD of the other side. this makes the denominators of both sides the same, so if they are equal, the numerators should end up being the same as well.so is there really a way to know what to multiply by?
i told you to do both sides separately.
say, "Consider the left hand side" ...and then do your stuff
then say, "Consider the right hand side" ... and do your stuff
then say, "We see that the right hand side simplifies to the same expression as the left hand side does. thus, the equation is true"
anyway, what's the hold up? expand the brackets and simplify
$\displaystyle \frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \frac{{\left( {\sqrt 6 - \sqrt 3 } \right)\left( {\sqrt 6 + \sqrt 3 } \right)}}
{{\sqrt 6 - \sqrt 3 }} + \frac{{\left( {\sqrt 7 - \sqrt 3 } \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}
{{\sqrt 7 + \sqrt 3 }}.$
So
$\displaystyle \frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \sqrt 7 + \sqrt 6 .$
Now multiply the last term by $\displaystyle \frac{{\sqrt 7 - \sqrt 6 }}
{{\sqrt 7 - \sqrt 6 }}$ and the rest follows.