What is (e^{iz}-e^{-iz})^3 ?

The solution to (e^{iz}-e^{-iz})^3 is -8isin^3(z)

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simplify (e^{iz}-e^{-iz})^3

simplify

(e^{i z} - e^{- i z})^{3}

- 8 i sin^{3} (z)

Solution steps

(e^{i z} - e^{- i z})^{3}

e^{i z} = cos (z) + i sin (z)

e^{- i z} = cos (z) - i sin (z)

= (cos (z) + i sin (z) - (cos (z) - i sin (z)))^{3}

Expand

cos (z) + sin (z) i - (cos (z) - i sin (z)) : 2 i sin (z)

= (2 i sin (z))^{3}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{3} i^{3} sin^{3} (z)

2^{3} = 8

= 8 i^{3} sin^{3} (z)

i^{3} = - i

= 8 (- i) sin^{3} (z)

Remove parentheses:

(- a) = - a

= - 8 i sin^{3} (z)

(e^{2 t} + 2 e^{- t})^{2}

\frac{( x ^{2} + 1 )}{x ( x + 1 ) ( x + 2 )}

\frac{x ^{3}}{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}

\frac{1}{a s ^{2}} - \frac{e ^{- a s}}{s ( 1 - e ^{- a s} )}

\frac{( a + b - 2 ab ) ( a + b )}{a - b}