What is expand ((x-1)(2x-1)(4x-1))/8 ?

The solution to expand ((x-1)(2x-1)(4x-1))/8 is x^3-(7x^2)/4+(7x)/8-1/8

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expand ((x-1)(2x-1)(4x-1))/8

expand

\frac{( x - 1 ) ( 2 x - 1 ) ( 4 x - 1 )}{8}

x^{3} - \frac{7 x ^{2}}{4} + \frac{7 x}{8} - \frac{1}{8}

Solution steps

\frac{( x - 1 ) ( 2 x - 1 ) ( 4 x - 1 )}{8}

Expand

(x - 1) (2 x - 1) (4 x - 1) : 8 x^{3} - 14 x^{2} + 7 x - 1

= \frac{8 x ^{3} - 14 x ^{2} + 7 x - 1}{8}

Apply the fraction rule:

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

\frac{8 x ^{3} - 14 x ^{2} + 7 x - 1}{8} = \frac{8 x ^{3}}{8} - \frac{14 x ^{2}}{8} + \frac{7 x}{8} - \frac{1}{8}

= \frac{8 x ^{3}}{8} - \frac{14 x ^{2}}{8} + \frac{7 x}{8} - \frac{1}{8}

Divide the numbers:

\frac{8}{8} = 1

= x^{3} - \frac{14 x ^{2}}{8} + \frac{7 x}{8} - \frac{1}{8}

Cancel

\frac{14 x ^{2}}{8} : \frac{7 x ^{2}}{4}

= x^{3} - \frac{7 x ^{2}}{4} + \frac{7 x}{8} - \frac{1}{8}

- \frac{s}{( s - a ) ( s ^{2} + 81 )}

\frac{1}{( 2 + π ) ^{10}}

48 (\frac{5}{6} x + 2 = \frac{7}{8})

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

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

What is expand ((x-1)(2x-1)(4x-1))/8 ?
The solution to expand ((x-1)(2x-1)(4x-1))/8 is x^3-(7x^2)/4+(7x)/8-1/8