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

The solution to expand ((2x-3)^3)/8 is x^3-(9x^2)/2+(27x)/4-27/8

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

expand

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

x^{3} - \frac{9 x ^{2}}{2} + \frac{27 x}{4} - \frac{27}{8}

Solution steps

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

(2 x - 3)^{3} = 8 x^{3} - 36 x^{2} + 54 x - 27

= \frac{8 x ^{3} - 36 x ^{2} + 54 x - 27}{8}

Apply the fraction rule:

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

\frac{8 x ^{3} - 36 x ^{2} + 54 x - 27}{8} = \frac{8 x ^{3}}{8} - \frac{36 x ^{2}}{8} + \frac{54 x}{8} - \frac{27}{8}

= \frac{8 x ^{3}}{8} - \frac{36 x ^{2}}{8} + \frac{54 x}{8} - \frac{27}{8}

Divide the numbers:

\frac{8}{8} = 1

= x^{3} - \frac{36 x ^{2}}{8} + \frac{54 x}{8} - \frac{27}{8}

Cancel

\frac{36 x ^{2}}{8} : \frac{9 x ^{2}}{2}

= x^{3} - \frac{9 x ^{2}}{2} + \frac{54 x}{8} - \frac{27}{8}

Cancel

\frac{54 x}{8} : \frac{27 x}{4}

= x^{3} - \frac{9 x ^{2}}{2} + \frac{27 x}{4} - \frac{27}{8}

\frac{a ( 2 x + 1 ) + 2}{a ( - x w - 1 + w )}

\frac{( 2 x - 3 ) ^{3}}{6}

\frac{d}{d x} ((z (x, y)) (x^{2} + 2 y^{2} - 6 x - 8 y + 20))

\frac{p x + q}{( x - a ) ( x - b )}

\frac{7 x - 10}{( x - 4 ) ( 2 x + 1 )}

What is expand ((2x-3)^3)/8 ?
The solution to expand ((2x-3)^3)/8 is x^3-(9x^2)/2+(27x)/4-27/8