What is expand 1/((s^2+1)(s-3)(s-6)) ?

The solution to expand 1/((s^2+1)(s-3)(s-6)) is 1/(s^4-9s^3+19s^2-9s+18)

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expand 1/((s^2+1)(s-3)(s-6))

expand

\frac{1}{( s ^{2} + 1 ) ( s - 3 ) ( s - 6 )}

\frac{1}{s ^{4} - 9 s ^{3} + 19 s ^{2} - 9 s + 18}

Solution steps

\frac{1}{( s ^{2} + 1 ) ( s - 3 ) ( s - 6 )}

Expand

(s^{2} + 1) (s - 3) (s - 6) : s^{4} - 9 s^{3} + 19 s^{2} - 9 s + 18

= \frac{1}{s ^{4} - 9 s ^{3} + 19 s ^{2} - 9 s + 18}

(2 + - 9)^{3}

- \frac{ln ( x - 3 )}{ln ( e )} + e^{l n (- 2)}

\frac{5 x ^{2} - 2}{x ^{3} ( x + 1 ) ^{2}}

8 \cdot 1 \frac{4}{5} + 2.4 \cdot \frac{1}{4} - \frac{1}{5} (0.02 + \frac{3}{100})

1 + (e^{- x^{2}} - 2 e^{- x^{2}} x^{2})^{2}

What is expand 1/((s^2+1)(s-3)(s-6)) ?
The solution to expand 1/((s^2+1)(s-3)(s-6)) is 1/(s^4-9s^3+19s^2-9s+18)