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}

Divide the numbers:

\frac{3}{3} = 1

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

(1 - u^{2})^{2} u^{2} - d u

(1 + 1 + 1 = 0,) (1 + 2 + 3 = 0)

52 (x^{'} + h (x))^{2}

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

A (3, - 2) B (3, - 6)

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