What is expand ((s-1)^3)/(s^4) ?

The solution to expand ((s-1)^3)/(s^4) is 1/s-3/(s^2)+3/(s^3)-1/(s^4)

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expand ((s-1)^3)/(s^4)

expand

\frac{( s - 1 ) ^{3}}{s ^{4}}

\frac{1}{s} - \frac{3}{s ^{2}} + \frac{3}{s ^{3}} - \frac{1}{s ^{4}}

Solution steps

\frac{( s - 1 ) ^{3}}{s ^{4}}

(s - 1)^{3} = s^{3} - 3 s^{2} + 3 s - 1

= \frac{s ^{3} - 3 s ^{2} + 3 s - 1}{s ^{4}}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{s ^{3} - 3 s ^{2} + 3 s - 1}{s ^{4}} = \frac{s ^{3}}{s ^{4}} - \frac{3 s ^{2}}{s ^{4}} + \frac{3 s}{s ^{4}} - \frac{1}{s ^{4}}

= \frac{s ^{3}}{s ^{4}} - \frac{3 s ^{2}}{s ^{4}} + \frac{3 s}{s ^{4}} - \frac{1}{s ^{4}}

Cancel

\frac{s ^{3}}{s ^{4}} : \frac{1}{s}

= \frac{1}{s} - \frac{3 s ^{2}}{s ^{4}} + \frac{3 s}{s ^{4}} - \frac{1}{s ^{4}}

Cancel

\frac{3 s ^{2}}{s ^{4}} : \frac{3}{s ^{2}}

= \frac{1}{s} - \frac{3}{s ^{2}} + \frac{3 s}{s ^{4}} - \frac{1}{s ^{4}}

Cancel

\frac{3 s}{s ^{4}} : \frac{3}{s ^{3}}

= \frac{1}{s} - \frac{3}{s ^{2}} + \frac{3}{s ^{3}} - \frac{1}{s ^{4}}

\frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

(t - a) H (t - a)

(4 x^{3} + \frac{6}{x ^{3}}) (x^{\frac{3}{2}} - 4 x)

(x - 1) (2 x + 5)

\frac{x + 2}{( x + 1 ) ( x - 2 )}

What is expand ((s-1)^3)/(s^4) ?
The solution to expand ((s-1)^3)/(s^4) is 1/s-3/(s^2)+3/(s^3)-1/(s^4)