What is expand ((2x-3)^3)/6 ?

The solution to expand ((2x-3)^3)/6 is (4x^3)/3-6x^2+9x-9/2

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expand ((2x-3)^3)/6

expand

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

\frac{4 x ^{3}}{3} - 6 x^{2} + 9 x - \frac{9}{2}

Solution steps

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

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

= \frac{8 x ^{3} - 36 x ^{2} + 54 x - 27}{6}

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}{6} = \frac{8 x ^{3}}{6} - \frac{36 x ^{2}}{6} + \frac{54 x}{6} - \frac{27}{6}

= \frac{8 x ^{3}}{6} - \frac{36 x ^{2}}{6} + \frac{54 x}{6} - \frac{27}{6}

\frac{8 x ^{3}}{6} = \frac{4 x ^{3}}{3}

\frac{36 x ^{2}}{6} = 6 x^{2}

\frac{54 x}{6} = 9 x

\frac{27}{6} = \frac{9}{2}

= \frac{4 x ^{3}}{3} - 6 x^{2} + 9 x - \frac{9}{2}

\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 )}

v (0, - 5) f (0, 4)

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

What is expand ((2x-3)^3)/6 ?
The solution to expand ((2x-3)^3)/6 is (4x^3)/3-6x^2+9x-9/2