What is expand-(2(-x+1)^3)/3 ?

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

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expand-(2(-x+1)^3)/3

expand

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

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

Solution steps

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

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

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

Expand

2 (- x^{3} + 3 x^{2} - 3 x + 1) : - 2 x^{3} + 6 x^{2} - 6 x + 2

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

Apply the fraction rule:

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

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

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

Remove parentheses:

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

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

Divide the numbers:

\frac{6}{3} = 2

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

\frac{( 3 sin ( θ ) + 3 ) ^{2}}{2}

\frac{1}{2 u ^{3}} (u \cdot \frac{1}{2} x - v + u^{3} = 0)

(sinh (x)) (coth^{2} (x) + csch^{3} (x)) d x

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

t^{2} (a t + b) e^{t}

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