Math Problem Solver

$$\frac{4 p}{{p}^{2} - p + 1} = \frac{4 \cdot \left(3 + \frac{1}{p}\right)}{\left(3 + \frac{1}{p}\right)^{2} - 2 - \frac{1}{p}}$$

$$\begin{aligned}\frac{4 p}{p^{2} - p + 1}&=4 \cdot \left(3 + \frac{1}{p}\right) \left(1 - 3 + \frac{1}{p} + \left(3 + \frac{1}{p}\right)^{2}\right)^{-1}\\&=\frac{4 \cdot \left(\frac{1}{p} + 3\right)}{\left(\left(\frac{1}{p} + 3\right)^{2} + 1 - \frac{1}{p} + 3\right)}\\&=\frac{4 \cdot \left(\frac{1}{p} + 3\right)}{\frac{1}{p^{2}} + \frac{6}{p} + 9 + 1 - \frac{1}{p} + 3}\\&=\frac{4 \cdot \left(\frac{1}{p} + 3\right)}{\frac{- 3 p - 1}{p} + \frac{1}{p^{2}} + \frac{6}{p} + 9 + 1}\\&=\frac{4 \cdot \left(\frac{1}{p} + 3\right)}{\frac{1}{p^{2}} + \frac{6}{p} + \frac{- 3 p - 1}{p} + 10}\\&=\frac{4 \cdot \left(\frac{1}{p} + 3\right)}{\frac{1}{p^{2}} + \frac{5}{p} + 7}\\&=\frac{4 \cdot \left(3 + \frac{1}{p}\right)}{\left(3 + \frac{1}{p}\right)^{2} - 2 - \frac{1}{p}}\end{aligned}$$