Question

simplify: [ k(1-p)^2(k-1)/ p^2) ] - [p^2 *kp(1-p) ]

Answer 1

We have
\[ k(1-p)^2 \frac{(k-1)}{p^2}- p^2\times kp(1-p)\]
Let's expand \((1-p)^2\) as \((1+p^2-2p)\)
now we'll have
\[ k(1+p^2-2p) \frac{(k-1)}{p^2}- p^2\times kp(1-p)\]
now let's expand the first term

\[ \frac{k (k-1)}{p^2} + \frac{k(k-1)p^2}{p^2} -\frac{k(k-1)(2p)}{p^2}-p^2(kp)(1-p)\]

Now cancelling the common terms
\[ \frac{k (k-1)}{p^2} + \frac{k(k-1)\cancel{p^2}}{\cancel{p^2}} -\frac{k(k-1)(2\cancel{p})}{p^\cancel{2}}-p^2(kp)(1-p)\]
we'll get now
\[ \frac{k (k-1)}{p^2} + k(k-1) -\frac{k(k-1)(2)}{p}-p^2(kp)(1-p)\]
Let's
take the k out of each term
\[k (\frac{(k-1)}{p^2}+(k-1)-\frac{2(k-1)}{p}- p^3 (1-p))\]
I think this is the answer, do you have the answer of this?

Answer 2

no the best i got to was:
(k-kp)^2 (k-1) /p^2 - p^2* (kp-kp^2)p^2/p^2