What is expand (s+1)(s+1+j^2pi)(s+1-j^2pi) ?

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

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expand (s+1)(s+1+j^2pi)(s+1-j^2pi)

expand

(s + 1) (s + 1 + j^{2} π) (s + 1 - j^{2} π)

s^{3} + 3 s^{2} - π^{2} j^{4} s + 3 s - π^{2} j^{4} + 1

Solution steps

(s + 1) (s + 1 + j^{2} π) (s + 1 - j^{2} π)

= (s + 1) (s + 1 + π j^{2}) (s + 1 - π j^{2})

Expand

(s + 1) (s + 1 + j^{2} π) : π j^{2} s + s^{2} + 2 s + π j^{2} + 1

= (π j^{2} s + s^{2} + 2 s + π j^{2} + 1) (s + 1 - j^{2} π)

Expand

(π j^{2} s + s^{2} + 2 s + π j^{2} + 1) (s + 1 - j^{2} π) : s^{3} + 3 s^{2} - π^{2} j^{4} s + 3 s - π^{2} j^{4} + 1

= s^{3} + 3 s^{2} - π^{2} j^{4} s + 3 s - π^{2} j^{4} + 1

\frac{x ( x + 3 )}{x - 4}

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

(z - \frac{5}{3}) (z - \frac{2}{3})

\frac{5 x ^{3} - 6 x ^{2} + 7 x - 8}{x ( x + 4 ) ( x - 4 ) ^{2}}

π (- 4 ln (2) + 4) - \frac{3 π ( - 4 ln ( 2 ) + 4 )}{4}

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