Question

solve this problem guys...

Answer 1

As numerator has greater power than denominator, we have to reduce it
[tex] \frac{n^4}{n^3 - 1} = \frac{n^4-n+n}{n^3-1} = \frac{n(n^3-1)+n}{n^3-1} \\ = n + \frac{n}{n^3-1} [/tex]
now solve by partial fraction
we know that 
$n^3 - 1 = (n-1) (n^2+n+1)$
so
[tex] \frac{n}{n^3-1} = \frac{A}{n-1} + \frac{Bn+C}{n^2+n+1} \\ n = A(n^2+n+1) + (Bn+C)(n-1)[/tex]
put n = 1
$1 = 3A \\ A = \frac{1}{3}$
now
$0*n^2 + n = \frac{n^2+n+1}{3} + (Bn^2+Cn-Bn-C)$
$0*n^2 + n = \frac{(3B+1)n^2+(3C-3B+1)n+(1-3C)}{3}$
By comparing $n^2$ on both sides
$B = - \frac{1}{3}$
By comparing constant term on both sides
$C = \frac{1}{3}$
Finally
$\frac{n^4}{n-1} = n + \frac{1}{3(n-1)} + \frac{1-n}{3(n^2+n+1)}$