What is expand ((2x-1)/(2x))^3 ?

The solution to expand ((2x-1)/(2x))^3 is 1-3/(2x)+3/(4x^2)-1/(8x^3)

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expand ((2x-1)/(2x))^3

expand

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

1 - \frac{3}{2 x} + \frac{3}{4 x ^{2}} - \frac{1}{8 x ^{3}}

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply exponent rule:

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

(2 x)^{3} = 2^{3} x^{3}

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

(2 x - 1)^{3} = 8 x^{3} - 12 x^{2} + 6 x - 1

= \frac{8 x ^{3} - 12 x ^{2} + 6 x - 1}{2 ^{3} x ^{3}}

2^{3} = 8

= \frac{8 x ^{3} - 12 x ^{2} + 6 x - 1}{8 x ^{3}}

Apply the fraction rule:

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

\frac{8 x ^{3} - 12 x ^{2} + 6 x - 1}{8 x ^{3}} = \frac{8 x ^{3}}{8 x ^{3}} - \frac{12 x ^{2}}{8 x ^{3}} + \frac{6 x}{8 x ^{3}} - \frac{1}{8 x ^{3}}

= \frac{8 x ^{3}}{8 x ^{3}} - \frac{12 x ^{2}}{8 x ^{3}} + \frac{6 x}{8 x ^{3}} - \frac{1}{8 x ^{3}}

Apply rule

\frac{a}{a} = 1

\frac{8 x ^{3}}{8 x ^{3}} = 1

= 1 - \frac{12 x ^{2}}{8 x ^{3}} + \frac{6 x}{8 x ^{3}} - \frac{1}{8 x ^{3}}

Cancel

\frac{12 x ^{2}}{8 x ^{3}} : \frac{3}{2 x}

= 1 - \frac{3}{2 x} + \frac{6 x}{8 x ^{3}} - \frac{1}{8 x ^{3}}

Cancel

\frac{6 x}{8 x ^{3}} : \frac{3}{4 x ^{2}}

= 1 - \frac{3}{2 x} + \frac{3}{4 x ^{2}} - \frac{1}{8 x ^{3}}

(3 x^{2} - \frac{1}{x ^{3}})^{10}

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

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

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

(7 - t) (7 + t)

What is expand ((2x-1)/(2x))^3 ?
The solution to expand ((2x-1)/(2x))^3 is 1-3/(2x)+3/(4x^2)-1/(8x^3)