What is expand ((3x+4)^3)/9 ?

The solution to expand ((3x+4)^3)/9 is 3x^3+12x^2+16x+64/9

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expand ((3x+4)^3)/9

expand

\frac{( 3 x + 4 ) ^{3}}{9}

3 x^{3} + 12 x^{2} + 16 x + \frac{64}{9}

Solution steps

\frac{( 3 x + 4 ) ^{3}}{9}

(3 x + 4)^{3} = 27 x^{3} + 108 x^{2} + 144 x + 64

= \frac{27 x ^{3} + 108 x ^{2} + 144 x + 64}{9}

Apply the fraction rule:

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

\frac{27 x ^{3} + 108 x ^{2} + 144 x + 64}{9} = \frac{27 x ^{3}}{9} + \frac{108 x ^{2}}{9} + \frac{144 x}{9} + \frac{64}{9}

= \frac{27 x ^{3}}{9} + \frac{108 x ^{2}}{9} + \frac{144 x}{9} + \frac{64}{9}

Divide the numbers:

\frac{27}{9} = 3

= 3 x^{3} + \frac{108 x ^{2}}{9} + \frac{144 x}{9} + \frac{64}{9}

Divide the numbers:

\frac{108}{9} = 12

= 3 x^{3} + 12 x^{2} + \frac{144 x}{9} + \frac{64}{9}

Divide the numbers:

\frac{144}{9} = 16

= 3 x^{3} + 12 x^{2} + 16 x + \frac{64}{9}

U (t - 4) (2 - t^{2}) + 3 e^{- π t} + sin (5 t) - t^{3}

9 (π + 1)

(A + B) (C + B) (D^{^{'}} + B) (A C D^{^{'}} + E)

(2 x + 3) d x + (3 x + 2 y) d y

\frac{5}{( x + 2 ) ^{2}}

What is expand ((3x+4)^3)/9 ?
The solution to expand ((3x+4)^3)/9 is 3x^3+12x^2+16x+64/9