# How would I apply partial fraction expansion to this expression?

$$\displaystyle\frac{1}{X\bigg(1-\dfrac{X}{Y}\bigg)\bigg(\dfrac{X}{Z}-1\bigg)}$$

I want it in the form $$\frac{A}{X} + \frac{B}{(1-\dfrac{X}{Y})}+\frac{C}{(\dfrac{X}{Z}-1)}$$where I am required to find the values for A, B and C. I have found $$B=C$$ but I am unsure if that is correct

I have tried simplifying it so that $$X=x, X/Y = y$$ and $$X/Z=z$$ but that has not helped.

• What are $X, Y, Z$? Are they all variables? Or are $Y, Z$ constants while $X$ is a variable? – kkc Mar 24 at 21:31

Given that $$\frac1{X(1-\frac{X}{Y})(\frac{X}{Z}-1)} = \frac{A}{X}+\frac{B}{1-\frac{X}{Y}}+\frac{C}{\frac{X}{Z}-1}$$ We'll multiply everything by the denominators here: $$1 = A\cdot(1-\frac{X}{Y})(\frac{X}{Z}-1) + B\cdot(X)(\frac{X}{Z}-1) + C\cdot(1-\frac{X}{Y})(X)$$ We want this statement to be true when $$\frac{X}{Y}=1$$. That's not necessarily saying that $$\frac{X}{Y}$$ will always be equal to one. We're saying, if it so happens that it does equal one, we need it to be true. So... Plug it in! And everything that has a $$(1-\frac{X}{Y})$$ term will cancel out! $$1 = B\cdot(X)(\frac{X}{Z}-1)$$ $$B=\frac1{(X)(\frac{X}{Z}-1)}$$ We also want that statment to be true if and when $$X=0$$. Using the same procedure... $$A=\frac1{(1-\frac{X}{Y})(\frac{X}{Z}-1)}$$ And with the same logic... $$C=\frac1{X(1-\frac{X}{Y})}$$
• You aren't making the substitutions. If $X=0,$ you get $A=-1,$ for example. The formulas for $A,B,C$ must be in terms of $Y$ and $Z$ only. – saulspatz Mar 24 at 21:44