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

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

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

expand

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

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

Solution steps

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

(2 - x)^{3} = 8 - 12 x + 6 x^{2} - x^{3}

= \frac{x ( - x ^{3} + 6 x ^{2} - 12 x + 8 )}{3}

Expand

x (8 - 12 x + 6 x^{2} - x^{3}) : 8 x - 12 x^{2} + 6 x^{3} - x^{4}

= \frac{8 x - 12 x ^{2} + 6 x ^{3} - x ^{4}}{3}

Apply the fraction rule:

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

\frac{8 x - 12 x ^{2} + 6 x ^{3} - x ^{4}}{3} = \frac{8 x}{3} - \frac{12 x ^{2}}{3} + \frac{6 x ^{3}}{3} - \frac{x ^{4}}{3}

= \frac{8 x}{3} - \frac{12 x ^{2}}{3} + \frac{6 x ^{3}}{3} - \frac{x ^{4}}{3}

Divide the numbers:

\frac{12}{3} = 4

= \frac{8 x}{3} - 4 x^{2} + \frac{6 x ^{3}}{3} - \frac{x ^{4}}{3}

Divide the numbers:

\frac{6}{3} = 2

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

Rewrite in standard form

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

(t cos (u), t sin (u), u) \cdot (sin (u), - cos (u), t)

(2 a + b) (2 a - b)

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

(3 - 4 i) (3 + 4 i)

\frac{1}{5} (15 + 10 k)

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