+50 Show that
(p3+1)/p2-(2p-6)/p = ap2+b/p+c/p2
where a, b and c are constants to be found
Expand the following:
(p^3+1)/p^2-(2 p-6)/p = c/p^2+b/p+a p^2
(2 p-6)/p = (2 p)/p-(6)/p:
(p^3+1)/p^2-(2 p)/p-(6)/p = c/p^2+b/p+a p^2
(2 p)/p = p/p×2 = 2:
-(2-(6)/p)+(p^3+1)/p^2 = c/p^2+b/p+a p^2
-(2-(6)/p) = -(-6)/p-2:
-(-6)/p-2+(p^3+1)/p^2 = c/p^2+b/p+a p^2
-(-6) = 6:
-2+6/p+(p^3+1)/p^2 = c/p^2+b/p+a p^2
(p^3+1)/(p^2) = p^3/p^2+1/p^2:
-2+p^3/p^2+1/p^2+6/p = c/p^2+b/p+a p^2
Combine powers. p^3/p^2 = p^(3-2):
p^3-2-2+6/p+1/p^2 = c/p^2+b/p+a p^2
3-2 = 1:
Answer: |p-2+6/p+1/p^2 = c/p^2+b/p+a p^2
