What is (l^{-1}1)/((s+a)(s-b)) ?

The solution to (l^{-1}1)/((s+a)(s-b)) is 1/(l(s+a)(s-b))

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simplify (l^{-1}1)/((s+a)(s-b))

simplify

\frac{l ^{- 1} 1}{( s + a ) ( s - b )}

\frac{1}{l ( s + a ) ( s - b )}

Solution steps

\frac{l ^{- 1} \cdot 1}{( s + a ) ( s - b )}

Apply rule:

a \cdot 1 = a

= \frac{l ^{- 1}}{( s + a ) ( s - b )}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

l^{- 1} = \frac{1}{l}

= \frac{\frac{1}{l}}{( s + a ) ( s - b )}

Apply the fraction rule:

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

= \frac{1}{l ( s + a ) ( s - b )}

216 - 108 b + 18 b^{2} - b^{3}

(6 x + y) (2 x - 5 y)

b (- 32) a (23) c (1 - 4)

\frac{( a + b ) ( a ^{3} - b ^{3} )}{( a ^{2} - b ^{2} )}

\frac{( 9 ^{n + 2} + 10 \cdot 9 ^{n} )}{( 9 ^{n + 1} \cdot 11 - 8 \cdot 9 ^{n} )}

What is (l^{-1}1)/((s+a)(s-b)) ?
The solution to (l^{-1}1)/((s+a)(s-b)) is 1/(l(s+a)(s-b))