What is expand-(x^2(2x-9))/6 ?

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

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expand-(x^2(2x-9))/6

expand

- \frac{x ^{2} ( 2 x - 9 )}{6}

- \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2}

Solution steps

- \frac{x ^{2} ( 2 x - 9 )}{6}

Expand

x^{2} (2 x - 9) : 2 x^{3} - 9 x^{2}

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

Apply the fraction rule:

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

\frac{2 x ^{3} - 9 x ^{2}}{6} = - (\frac{2 x ^{3}}{6}) - (- \frac{9 x ^{2}}{6})

= - (\frac{2 x ^{3}}{6}) - (- \frac{9 x ^{2}}{6})

Remove parentheses:

(a) = a, - (- a) = a

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

Cancel

\frac{2 x ^{3}}{6} : \frac{x ^{3}}{3}

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

Cancel

\frac{9 x ^{2}}{6} : \frac{3 x ^{2}}{2}

= - \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2}

(cos (x) + cos (y))^{2}

\frac{x ^{m} ( x ^{m} + 1 )}{2}

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

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

x^{2} \cdot (\frac{1}{2 x ^{\frac{1}{2}}} + 1)

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