What is expand (2^'s+3)/((s-1)^3(s^3+1)) ?

The solution to expand (2^'s+3)/((s-1)^3(s^3+1)) is (2^'s+3)/(s^6-3s^5+3s^4-3s^2+3s-1)

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expand (2^'s+3)/((s-1)^3(s^3+1))

expand

\frac{2 ^{^{'}} s + 3}{( s - 1 ) ^{3} ( s ^{3} + 1 )}

\frac{2 ^{'} s + 3}{s ^{6} - 3 s ^{5} + 3 s ^{4} - 3 s ^{2} + 3 s - 1}

Solution steps

\frac{2 ^{^{'}} s + 3}{( s - 1 ) ^{3} ( s ^{3} + 1 )}

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

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

Expand

(s^{3} - 3 s^{2} + 3 s - 1) (s^{3} + 1) : s^{6} - 3 s^{5} + 3 s^{4} - 3 s^{2} + 3 s - 1

= \frac{2 ^{^{'}} s + 3}{s ^{6} - 3 s ^{5} + 3 s ^{4} - 3 s ^{2} + 3 s - 1}

(x^{2} + 1)^{2} (2 x) d x

(- 8 t^{3} + 68 t^{2} - 204 t + 207) \cdot (- 1, + 2)

\frac{( a y + b x ) ( x a + y b )}{( a y - b x ) ( x a - y b )}

\frac{4 x}{ln ( 10 ) ( x ^{2} + 1 )}

\frac{d}{d x} ((z) (4 z^{3} + 3 x z^{2} - x y z - 2 x y^{2} + 7))

What is expand (2^'s+3)/((s-1)^3(s^3+1)) ?
The solution to expand (2^'s+3)/((s-1)^3(s^3+1)) is (2^'s+3)/(s^6-3s^5+3s^4-3s^2+3s-1)