What is expand 3x(((3-x))/9)^2 ?

The solution to expand 3x(((3-x))/9)^2 is (x^3}{27}-\frac{2x^2)/9+x/3

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expand 3x(((3-x))/9)^2

expand

3 x (\frac{( 3 - x )}{9})^{2}

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

Solution steps

3 x (\frac{3 - x}{9})^{2}

(\frac{3 - x}{9})^{2} = \frac{9 - 6 x + x ^{2}}{9 ^{2}}

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

Multiply fractions:

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

= \frac{( 9 - 6 x + x ^{2} ) \cdot 3 x}{9 ^{2}}

Factor

9^{2} : 3^{4}

= \frac{3 x ( x ^{2} - 6 x + 9 )}{3 ^{4}}

Cancel the common factor:

3

= \frac{x ( x ^{2} - 6 x + 9 )}{3 ^{3}}

3^{3} = 27

= \frac{x ( x ^{2} - 6 x + 9 )}{27}

Expand

x (x^{2} - 6 x + 9) : x^{3} - 6 x^{2} + 9 x

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

Apply the fraction rule:

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

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

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

Cancel

\frac{6 x ^{2}}{27} : \frac{2 x ^{2}}{9}

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

Cancel

\frac{9 x}{27} : \frac{x}{3}

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

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

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

\frac{1}{2} (x - 2) (x - 6)

\frac{2 ( - r + 45 )}{5}

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

What is expand 3x(((3-x))/9)^2 ?
The solution to expand 3x(((3-x))/9)^2 is (x^3}{27}-\frac{2x^2)/9+x/3