What is expand-(x(-x+1)^3)/3 ?

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

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expand-(x(-x+1)^3)/3

expand

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

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

Solution steps

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

(- x + 1)^{3} = - x^{3} + 3 x^{2} - 3 x + 1

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

Expand

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

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

Apply the fraction rule:

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

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

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

Remove parentheses:

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

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

Divide the numbers:

\frac{3}{3} = 1

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

\frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x})

x y (x + y) z^{^{'}} + y

3 x (e^{x} - e^{- x})

(3 x^{2} y + 6 x) d x - (x^{3} + 3 y^{2}) d y

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