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

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

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

expand

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

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

Solution steps

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

Expand

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

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{8}{8} = 1

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

\frac{( e + 1 ) ( e - 1 )}{e ^{2} + 1}

\frac{b ^{2} ( 4 a ^{3} b ^{2} - 3 )}{a}

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

\frac{3 x ^{2}}{( x - 4 ) ^{2}}

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

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