What is expand 1/4 ((2-x)/3)^3 ?

The solution to expand 1/4 ((2-x)/3)^3 is -(x^3)/(108)+(x^2)/(18)-x/9+2/27

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expand 1/4 ((2-x)/3)^3

expand

\frac{1}{4} (\frac{2 - x}{3})^{3}

- \frac{x ^{3}}{108} + \frac{x ^{2}}{18} - \frac{x}{9} + \frac{2}{27}

Solution steps

\frac{1}{4} (\frac{2 - x}{3})^{3}

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

= \frac{1}{4} \cdot \frac{- x ^{3} + 6 x ^{2} - 12 x + 8}{3 ^{3}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot ( 8 - 12 x + 6 x ^{2} - x ^{3} )}{4 \cdot 3 ^{3}}

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

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

4 \cdot 3^{3} = 108

= \frac{8 - 12 x + 6 x ^{2} - x ^{3}}{108}

Apply the fraction rule:

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

\frac{8 - 12 x + 6 x ^{2} - x ^{3}}{108} = \frac{8}{108} - \frac{12 x}{108} + \frac{6 x ^{2}}{108} - \frac{x ^{3}}{108}

= \frac{8}{108} - \frac{12 x}{108} + \frac{6 x ^{2}}{108} - \frac{x ^{3}}{108}

Cancel

\frac{8}{108} : \frac{2}{27}

= \frac{2}{27} - \frac{12 x}{108} + \frac{6 x ^{2}}{108} - \frac{x ^{3}}{108}

Cancel

\frac{12 x}{108} : \frac{x}{9}

= \frac{2}{27} - \frac{x}{9} + \frac{6 x ^{2}}{108} - \frac{x ^{3}}{108}

Cancel

\frac{6 x ^{2}}{108} : \frac{x ^{2}}{18}

= \frac{2}{27} - \frac{x}{9} + \frac{x ^{2}}{18} - \frac{x ^{3}}{108}

Rewrite in standard form

= - \frac{x ^{3}}{108} + \frac{x ^{2}}{18} - \frac{x}{9} + \frac{2}{27}

(x - 2 a x + a)^{2}

(\frac{x}{3} - \frac{2 y}{5}) (\frac{x}{3} + \frac{2 y}{5})

\frac{x ^{2} - x - 1}{( x + 5 ) ( x ^{2} + 4 )}

4 (e^{- x} x - 2 e^{- x})

What is expand 1/4 ((2-x)/3)^3 ?
The solution to expand 1/4 ((2-x)/3)^3 is -(x^3)/(108)+(x^2)/(18)-x/9+2/27