What is (1-1/(jkr))(1+1/(jkr)-1/(k^2r^2)) ?

The solution to (1-1/(jkr))(1+1/(jkr)-1/(k^2r^2)) is ((jkr-1)(k^2jr^2+kr-j))/(k^3j^2r^3)

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simplify (1-1/(jkr))(1+1/(jkr)-1/(k^2r^2))

simplify

(1 - \frac{1}{jk r}) (1 + \frac{1}{jk r} - \frac{1}{k ^{2} r ^{2}})

\frac{( jk r - 1 ) ( k ^{2} j r ^{2} + k r - j )}{k ^{3} j ^{2} r ^{3}}

Solution steps

(1 - \frac{1}{jk r}) (1 + \frac{1}{jk r} - \frac{1}{k ^{2} r ^{2}})

Join

1 - \frac{1}{jk r} : \frac{jk r - 1}{jk r}

= \frac{jk r - 1}{jk r} (1 + \frac{1}{jk r} - \frac{1}{k ^{2} r ^{2}})

Apply the fraction rule:

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

= \frac{( jk r - 1 ) ( 1 + \frac{1}{jk r} - \frac{1}{k ^{2} r ^{2}} )}{jk r}

Simplify

(jk r - 1) (1 + \frac{1}{jk r} - \frac{1}{k ^{2} r ^{2}}) : \frac{( jk r - 1 ) ( k ^{2} j r ^{2} + k r - j )}{k ^{2} j r ^{2}}

= \frac{\frac{( jk r - 1 ) ( k ^{2} j r ^{2} + k r - j )}{k ^{2} j r ^{2}}}{jk r}

Apply the fraction rule:

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

= \frac{( jk r - 1 ) ( k ^{2} j r ^{2} + k r - j )}{k ^{2} j r ^{2} jk r}

Simplify

k^{2} j r^{2} jk r : k^{3} j^{2} r^{3}

= \frac{( jk r - 1 ) ( k ^{2} j r ^{2} + k r - j )}{k ^{3} j ^{2} r ^{3}}

(e^{t} cos (t) - e^{t} sin (t)) \cdot (2 e^{t} cos (t))

\frac{( 5 x - 22 ) ( x - 1 )}{x + 5}

\frac{\partial}{\partial x} ((y) (x^{3} y - x - 5 y - 10))

(x^{2} - 5)^{8} \cdot 2 x d x

\frac{( 4 - 3 i ) ^{6}}{6}

What is (1-1/(jkr))(1+1/(jkr)-1/(k^2r^2)) ?
The solution to (1-1/(jkr))(1+1/(jkr)-1/(k^2r^2)) is ((jkr-1)(k^2jr^2+kr-j))/(k^3j^2r^3)