Factor the denominator:
#x^6-x^4 - x^4(x^2-1) = x^4(x+1)(x-1)#
Perhaps we know that #x^5+1# can be factored, or perhaps we notice that #-1# is a zero of the numerator, so #x-1# is a factor.
In any case, we can write:
#(x^5 + 1)/(x^6 - x^4) = ((x+1)(x^4-x^3+x^2-x+1))/(x^4(x+1)(x-1))#
So
#(x^5 + 1)/(x^6 - x^4) = (x^4-x^3+x^2-x+1)/(x^4(x-1))#
#A/x+B/x^2+C/x^3+D/x^4+E/(x-1) = (x^4-x^3+x^2-x+1)/(x^4(x-1))#
Leads to:
#(Ax^4-Ax^3+Bx^3-Bx^2+Cx^2-Cx+Dx-D+Ex^4)/(x^4(x-1))#
# = (x^4-x^3+x^2-x+1)/(x^4(x-1))#
So
#A+E = 1#
#-A+B=-1#
#-B+C = 1#
#-C+D=-1#
#-D = 1#
So #D=-1# and,working back through the list:
#C =0" "#, #B = -1" "#, #A = 0" "#, and #" "E = 1#.
But what if? What if we didn't notice that the fraction can be reduced?
It's OK, just a bit longer.
We would get:
#(x^5 + 1)/(x^6 - x^4) = (x^5+1)/(x^4(x+1)(x-1))#
#A/x+B/x^2+C/x^3+D/x^4+E/(x-1) +F/(x+1) = (x^5+1)/(x^4(x+1)(x-1))#
In this case we get one more variable (and one more equation).
When we solve, we'll get the same #A, B, C, D, "and " E# and we will get #F=0#
