I need to prove that: $$ \frac{1}{(x-a)(x-b)} = \frac{1}{(b-a)(x-b)}- \frac{1}{(b-a)(x-a)}, $$ and I must note that I need to go from the left expression to the right (because of the exercise).

So, I tried to solve $$ \frac{1}{(x-a)(x-b)} = \frac{C}{(x-a)} + \frac{D}{(x-b)}. $$

Unfortunately, from there, I am reaching a system that I can not solve. I get $C + D = 0$ and $Cb = -Da + 1$. From here I am not being able to make any progress. Any suggestions?

  • $\begingroup$ Hint: clear fractions and use the cover-up method. $\endgroup$ Apr 15 at 0:47
  • 3
    $\begingroup$ I think you're mixing up $C$ and $D$ with $A$ and $B$ and also $A$ and $B$ with $a$ and $b$. You've obtained two linear equations in the variables $C$ and $D$, and the usual methods for solving them should work here. (What would you do if $a$ and $b$ were numbers rather than letters?) $\endgroup$ Apr 15 at 0:50
  • $\begingroup$ yes, sorry, i forgot to put C and D instead of A and B, respectively. But still, i am getting to those two equations and i haven't been able to make any progress. $\endgroup$ Apr 15 at 0:54
  • $\begingroup$ @SeanRoberson i will look into that method, never heard of it $\endgroup$ Apr 15 at 0:55
  • $\begingroup$ Multiply both sides by $x-a$, then let $x=a$ to get $C$. Multiply both sides by $x-b$, then let $x=b$ to get $D$. $\endgroup$ Apr 15 at 0:59

2 Answers 2


Another possible approach (under the assumption that $a\neq b$): \begin{align*} \frac{1}{(x - a)(x - b)} & = \frac{1}{b - a}\frac{b - a}{(x - a)(x - b)}\\\\ & = \frac{1}{b - a}\frac{(x - a) - (x - b)}{(x - a)(x - b)}\\\\ & = \frac{1}{b - a}\left(\frac{1}{x - b} - \frac{1}{x - a}\right)\\\\ & = \frac{1}{(b - a)(x - b)} - \frac{1}{(b - a)(x - a)} \end{align*}


Lets continue from what you found:

$\begin{cases}C+D=0\Rightarrow C=-D\\Cb=-Da+1\Rightarrow D(a-b)=1\end{cases}$

From second equation we get:

$D=\frac1{a-b}=-\frac 1{b-a}\Rightarrow C=\frac 1{b-a}$

Putting C and D in relation you get:



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