Choose one of the two variables and collect together like powers removing common factors on the way. So, choosing p,
\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2}\\=p^{2}(1-2r+r^{2})-2p(1-r)+1 \\ =p^{2}(1-r)^{2}-2p(1-r)+1.\)
If you don't see how to simplyfy that directly, let p(1 - r) = X say, and you then have
\(\displaystyle X^{2}-2X+1 = (X-1)^{2}, \\ = \{p(1-r)-1\}^{2}.\)
Alternatively had you chosen r, you get
\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2} \\ =r^{2}p^{2}+2rp(1-p)+1-2p+p^{2}\\ = r^{2}p^{2}+2rp(1-p)+(1-p)^{2}.\)
This time, if needed, let rp = X and 1 - p = Y and you get
\(\displaystyle X^{2}+2XY+Y^{2}=(X+Y)^{2} \\=(rp+1-p)^{2}.\)
The two results differ by a negative sign within the brackets, but algebraically they're equivalent.
And btw, they are different from your given result, (which was incorrect).
