Answer
$a.\displaystyle \quad 1+\frac{A}{t}+\frac{B}{t^{2}}+\frac{C}{t^{3}}+\frac{D}{t+1}+\frac{Ex+F}{t^{2}-t+1}$
$b.\displaystyle \quad \frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^{2}+1}+\frac{Ex+F}{(x^{2}+1)^{2}}$
Work Step by Step
$a.$
Factor out $t^{3}$ in the denominator.
The numerator and denominator have the same degree. Use either long division or
$\displaystyle \frac{t^{6}+1}{t^{6}+t^{3}}=\frac{(t^{6}+t^{3})-t^{3}+1}{t^{6}+t^{3}}=1+\frac{-t^{3}+1}{t^{3}(t^{3}+1)}$
The second factor in the denominator is a sum of cubes,
$=1+\displaystyle \frac{-t^{3}+1}{t^{3}(t+1)(t^{2}-t+1)}$
Repeated factors,a linear factor, a quadratic factor.
$=1+\displaystyle \frac{A}{t}+\frac{B}{t^{2}}+\frac{C}{t^{3}}+\frac{D}{t+1}+\frac{Ex+F}{t^{2}-t+1}$
$b.$
Denominator:
$x^{2}-x=x(x-1)$
$x^{4}+2x^{2}+1=(x^{2}+1)^{2}$
$\displaystyle \frac{x^{5}+1}{(x^{2}-x)(x^{4}+2x^{2}+1)}=\frac{x^{5}+1}{x(x-1)(x^{2}+1)^{2}}$
Repeated quadratic factor.
$=\displaystyle \frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^{2}+1}+\frac{Ex+F}{(x^{2}+1)^{2}}$
