What is the inverse of (a(s^2-2a^2))/(s^4+4a^4) ?

The inverse of (a(s^2-2a^2))/(s^4+4a^4) is

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inverse of (a(s^2-2a^2))/(s^4+4a^4)

inverse

\frac{a ( s ^{2} - 2 a ^{2} )}{s ^{4} + 4 a ^{4}}

\frac{a + a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, - \frac{a + a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, \frac{a - a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, - \frac{a - a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}

Solution steps

y = \frac{a ( s ^{2} - 2 a ^{2} )}{s ^{4} + 4 a ^{4}}

Replace

s

with

y

s = \frac{a ( y ^{2} - 2 a ^{2} )}{y ^{4} + 4 a ^{4}}

Solve for

y, s = \frac{a ( y ^{2} - 2 a ^{2} )}{y ^{4} + 4 a ^{4}}

\frac{a + a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, - \frac{a + a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, \frac{a - a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}, - \frac{a - a ^{2} - 4 s ( 4 a ^{4} s + 2 a ^{3} )}{2 s}

f (x) = \frac{- 9 - 7 x}{8}

f (x) = \frac{9 x}{x - 3}

f (x) = - 9 x - 8 + 5, x \geq 8

\frac{1}{2} (e^{x} - e^{- x})

f (x) = a + \frac{1}{x - a}

What is the inverse of (a(s^2-2a^2))/(s^4+4a^4) ?
The inverse of (a(s^2-2a^2))/(s^4+4a^4) is