Question 125695
{{{6/i - 8/(8-i)}}}.


The first thing to do is rationalize the denominators, i.e. get the i out of each denominator and then you will be able to find a LCD to allow you to add the fractions.


The first fraction is easy, you just multiply by {{{i/i}}}, giving you {{{-6i/1}}}  (remember {{{i^2=-1}}})


The other fraction requires you to remember the factorization of the difference of two squares.  {{{a^2-b^2=(a+b)(a-b)}}}.  Using this, we could multiply the denominator of the second fraction, {{{8-i}}} by its conjugate {{{8+i}}} to obtain {{{8^2-i^2=64+1=65}}}, but to do that we also have to multiply the numerator by the same thing.  The result is that the second fraction looks like{{{-(8(8+i))/65}}}


Putting it all together we get {{{-(6i/1)-(8(8+i))/65}}}


65 is clearly the LCD, so


{{{(-390i-64-8i)/65=red((-64/65)-(398i/65))}}}