How to solve this partial fraction decomposition equation?

Question

$$\frac{1}{(x-a)(x-b)(x-c)} = \frac{A}{(x-a)} +\frac{B}{(x-b)} +\frac{C}{(x-c)}$$

a,b and c are constants.

I have tried three times by hand but I can't get the right result.

The result is $$\frac{1}{(x-a)(x-b)(x-c)} = \frac{1}{(a-b)(a-c)(x-a)} +\frac{1}{(b-a)(b-c)(x-b)} +\frac{1}{(c-a)(c-b)(x-c)}$$

Is there any easy and clear solution to this?

Answer 1

Hint

Starting with $$\frac{1}{(x-a)(x-b)(x-c)} = \frac{A}{(x-a)} +\frac{B}{(x-b)} +\frac{C}{(x-c)}$$ multiply by ${(x-a)(x-b)(x-c)}$ to get $$1=A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)$$ Now, make $$x=a\implies 1=A(a-b)(a-c)\implies A= ???$$ $$x=b\implies 1=B(b-a)(b-c)\implies B= ???$$ $$x=c\implies 1=C(c-a)(c-b)\implies C= ???$$

Answer 2

$$A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)=1$$ $$A(x^2-(b+c)x+bc)+B(x^2-(a+c)x-ac)+C(x^2-(a+b)x-ab)=1$$

Equate coefficients.

Coefficient of $x^2$ is 0 $$(A+B+C)x^2=0x^2$$ Thus $A+B+C=0$

You can form similar equations and then solve them by equating the coefficients of $x$ and $1$

Answer 3

$$\frac{1}{(x-a)(x-b)(x-c)} = \frac{A}{(x-a)} +\frac{B}{(x-b)} +\frac{C}{(x-c)}$$

Rearrange to give, $$A = \frac{1}{(x-b)(x-c)} - \frac{B(x-a)}{(x-b)} -\frac{C(x-a)}{(x-c)} $$

Put $x = a$ to get

$$A = \frac{1}{(a-b)(a-c)} $$

Similiarly,

$$B = \frac{1}{(b-a)(b-c)} $$

And

$$C = \frac{1}{(c-a)(c-b)} $$

Answer 4

The easy way is to follow the hint by @JackD'Aurizio, but let me spell it out in a little more detail. Start with $$\frac{1}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} +\frac{B}{x-b} +\frac{C}{x-c}$$ Multiply by $(x-a)$: $$\frac{1}{(x-b)(x-c)} = A + (x-a) \left( \frac{B}{x-b} + \frac{C}{x-c} \right)$$ Now set $x=a$, with the result $$\frac{1}{(a-b)(a-c)} = A$$ because all the stuff with a factor of $(x-a)$ vanishes.

$B$ and $C$ are obtained similarly. Voila!