What is (1-1/x)(1+1/(z^2))(1+1/z) ?

The solution to (1-1/x)(1+1/(z^2))(1+1/z) is ((x-1)(z^2+1)(z+1))/(z^3x)

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simplify (1-1/x)(1+1/(z^2))(1+1/z)

simplify

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

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

Solution steps

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

Join

1 - \frac{1}{x} : \frac{x - 1}{x}

= \frac{x - 1}{x} (1 + \frac{1}{z ^{2}}) (1 + \frac{1}{z})

Apply the fraction rule:

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

= \frac{( x - 1 ) ( 1 + \frac{1}{z ^{2}} ) ( 1 + \frac{1}{z} )}{x}

Simplify

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

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

Apply the fraction rule:

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

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

\frac{4 s + 2 s ^{2} + 8}{( s ^{2} + 4 ) ( s + 1 ) ^{2}}

\frac{6}{( s ^{'} + 7 s + 10 ) ( s ^{2} + 4 )}

(x + 1 - 2 i) (x + 1 + 2 i)

\frac{6 x ^{2} + 21 x + 11}{( x ^{2} + 3 ) ( x + 5 )}

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

What is (1-1/x)(1+1/(z^2))(1+1/z) ?
The solution to (1-1/x)(1+1/(z^2))(1+1/z) is ((x-1)(z^2+1)(z+1))/(z^3x)