What is expand ((3x+1)^3)/9 ?

The solution to expand ((3x+1)^3)/9 is 3x^3+3x^2+x+1/9

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expand ((3x+1)^3)/9

expand

\frac{( 3 x + 1 ) ^{3}}{9}

3 x^{3} + 3 x^{2} + x + \frac{1}{9}

Solution steps

\frac{( 3 x + 1 ) ^{3}}{9}

(3 x + 1)^{3} = 27 x^{3} + 27 x^{2} + 9 x + 1

= \frac{27 x ^{3} + 27 x ^{2} + 9 x + 1}{9}

Apply the fraction rule:

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

\frac{27 x ^{3} + 27 x ^{2} + 9 x + 1}{9} = \frac{27 x ^{3}}{9} + \frac{27 x ^{2}}{9} + \frac{9 x}{9} + \frac{1}{9}

= \frac{27 x ^{3}}{9} + \frac{27 x ^{2}}{9} + \frac{9 x}{9} + \frac{1}{9}

Divide the numbers:

\frac{27}{9} = 3

= 3 x^{3} + 3 x^{2} + \frac{9 x}{9} + \frac{1}{9}

Divide the numbers:

\frac{9}{9} = 1

= 3 x^{3} + 3 x^{2} + x + \frac{1}{9}

2 (\frac{5}{4}) (- s)

\frac{\frac{3}{4} ( 2 t sin ( 2 t ) + cos ( 2 t ) ) + C}{t}

3 (2 - 2)

(\frac{2}{7} - \frac{1}{x}) \cdot (\frac{5}{x} + \frac{1}{3})

4 ((\frac{19}{12})^{\frac{1}{24}} - 1)

What is expand ((3x+1)^3)/9 ?
The solution to expand ((3x+1)^3)/9 is 3x^3+3x^2+x+1/9