The question is to develop $$(x-\alpha)(x-\beta)(x-\gamma)\over(x-a)(x-b)$$ into partial fractions.
Someone challenged me to solve this question and said the answer is
$${x-(\alpha+\beta+\gamma-a-b)}+{(a-\alpha)(a-\beta)(a-\gamma)\over(x-a)(a-b)}+{(b-\alpha)(b-\beta)(b-\gamma)\over(x-b)(b-a)}$$
I tried and what I've done is
$$(x-\alpha)(x-\beta)(x-\gamma) = A(x-b)+B(x-a)$$
If $x=a$ then $$ A= \frac{(a-\alpha)(a-\beta)(a-\gamma)}{(a-b)}$$
If $x=b$ then $$B= \frac{(b-\alpha)(b-\beta)(b-\gamma)}{(b-a)}$$
Then the answer should be
$${(a-\alpha)(a-\beta)(a-\gamma)\over(x-a)(a-b)}+{(b-\alpha)(b-\beta)(b-\gamma)\over(x-b)(b-a)}$$
But this is not the right answer.
I dont know what's going wrong, I could use some help.
You have reached at $(x-\alpha)(x-\beta)(x-\gamma)$ = $A(x-b)+B(x-a)$
The left hand side is cubic whereas the right hand side is a linear polynomial.
As the degree of the numerator$(D_n)>$ that of the denominator$(D_d)$, in fact, $D_n-D_d=1$
using Partial Fraction Decomposition formula, $$\frac{(x-\alpha)(x-\beta)(x-\gamma)}{(x-a)(x-b)}=x^1+A+\dfrac B{x-a}+\dfrac C{x-b}$$
where $A,B,C$ are arbitrary constants.