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

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

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

expand

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

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

Solution steps

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{6}{3} = 2

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

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

e^{x} (1 + e^{- x})

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

X \cdot Y + λ (\frac{X}{4} + \frac{Y}{9} - 1)

\frac{1}{324} ((2 h - 33 y)^{4} - h^{4})

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