# tricky completing the square

$$\frac{4k_{0}p_{1}+p_{0}\left ( k_{0}^{2}-2k_{0}k_{1}+k_{1}^{2} \right )}{\left ( k_{0}+k_{1} \right )^{2}p_{0}}$$

I need to show that this is equal to $1$ but for my life I can't figure how to arrive at one despite multiple attempts at completing the square. This is an assignment in Quantum mechanics so obviously the focus is not completing the square. But I'd like to see how to arrive at one and the trick(s) involved.

• It is only true if the $4k_0 p_1$ was $4k_0 k_1 p_0$ in the numerator – lEm Apr 11 '16 at 9:15
• Maybe it was a typo? Because it depends on the value of $k_0 ,k_1, p_0, p_1$ specifically. It is obvious that they would not simply cancel out because there is a $p_1$ in the numerator but not in the denominator. – lEm Apr 11 '16 at 9:21
• What if $p_{0}=k_{0}$ $\space$ and $p_{1}=k_{1}$? Would anything change? – Mathematicing Apr 11 '16 at 9:29