What is expand ((2x-1)^3)/8 ?

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

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expand ((2x-1)^3)/8

expand

\frac{( 2 x - 1 ) ^{3}}{8}

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

Solution steps

\frac{( 2 x - 1 ) ^{3}}{8}

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

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

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

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

Divide the numbers:

\frac{8}{8} = 1

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

Cancel

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

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

Cancel

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

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

\frac{x ^{2} + y ^{2}}{( x + y ) ( x - y )}

(x + 6 - 9 x + 70)^{2}

x \cdot (1 - e^{- \frac{1}{x}})

(x + 5 - 3) (x^{2} + 9 + 5)

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