How to quickly solve partial fractions equation?

If your fraction is in form of $$ \frac {1}{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)...(t-\alpha_k)}$$ where the $\alpha s$ are different, there is an easy way to find the partial fraction decomposition.

Let $t=\alpha_i$ and cover $(t-\alpha_i)$ while evaluating the above fraction to get your $A_i$

For example $$ \frac{1}{(t-1)(t-3)(t+4)} = \frac {A_1}{(t-1)} + \frac{ A_2}{(t-3)} +\frac { A_3}{(t+4)}$$ Where $$A_1 = \frac {1}{(1-3)(1+4)}=\frac{-1}{10}$$ $$A_2 = \frac {1}{(3-1)(3+4)} = \frac{1}{14}$$

$$A_3=\frac {1}{(-4-1)(-4-3)}=\frac{1}{35}$$

Unless I am misunderstanding the question, by "root rotation" (when $\beta \neq \alpha$):

$$\frac{1}{(t + \alpha)(t + \beta)} = \frac{A}{t + \alpha} + \frac{B}{t + \beta}$$

$$1 = A(t + \beta) + B(t + \alpha)$$

Evaluating $-\beta$ for $t$:

$$1 = B(\alpha - \beta)$$

$$B = \frac{1}{\alpha - \beta}$$

Similarly, for $A$, sub in $-\alpha$:

$$1 = A(\beta - \alpha)$$

$$A = \frac{1}{\beta - \alpha}$$

Here's your answer for general $n$.

$\dfrac1{\prod_{k=1}^n (x-a_k)} =\sum_{k=1}^n \dfrac{b_k}{x-a_k} $.

Therefore $1 =\sum_{k=1}^n \dfrac{b_k\prod_{j=1}^n (x-a_j)}{x-a_k} =\sum_{k=1}^n b_k\prod_{j=1, j\ne k}^n (x-a_j) $.

Setting $x = a_i$ for each $i$, all the terms except the one with $k=i$ have the factor $a_i-a_i$, so $1 = b_i\prod_{j=1, j\ne i}^n (a_i-a_j) $ so that $b_i =\dfrac1{\prod_{j=1, j\ne i}^n (a_i-a_j)} $.

For $n=2$, $b_1 =\dfrac1{a_1-a_2} $, $b_2 =\dfrac1{a_2-a_1} $.

For $n=3$, $b_1 =\dfrac1{(a_1-a_2)(a_1-a_3)} $, $b_2 =\dfrac1{(a_2-a_1)(a_2-a_3)} $, $b_3 =\dfrac1{(a_3-a_1)(a_3-a_2)} $.