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

The solution to expand ((x^3+1)^3)/(9x^2) is (x^7)/9+(x^4)/3+x/3+1/(9x^2)

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expand ((x^3+1)^3)/(9x^2)

expand

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

\frac{x ^{7}}{9} + \frac{x ^{4}}{3} + \frac{x}{3} + \frac{1}{9 x ^{2}}

Solution steps

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

(x^{3} + 1)^{3} = x^{9} + 3 x^{6} + 3 x^{3} + 1

= \frac{x ^{9} + 3 x ^{6} + 3 x ^{3} + 1}{9 x ^{2}}

Apply the fraction rule:

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

\frac{x ^{9} + 3 x ^{6} + 3 x ^{3} + 1}{9 x ^{2}} = \frac{x ^{9}}{9 x ^{2}} + \frac{3 x ^{6}}{9 x ^{2}} + \frac{3 x ^{3}}{9 x ^{2}} + \frac{1}{9 x ^{2}}

= \frac{x ^{9}}{9 x ^{2}} + \frac{3 x ^{6}}{9 x ^{2}} + \frac{3 x ^{3}}{9 x ^{2}} + \frac{1}{9 x ^{2}}

Cancel

\frac{x ^{9}}{9 x ^{2}} : \frac{x ^{7}}{9}

= \frac{x ^{7}}{9} + \frac{3 x ^{6}}{9 x ^{2}} + \frac{3 x ^{3}}{9 x ^{2}} + \frac{1}{9 x ^{2}}

Cancel

\frac{3 x ^{6}}{9 x ^{2}} : \frac{x ^{4}}{3}

= \frac{x ^{7}}{9} + \frac{x ^{4}}{3} + \frac{3 x ^{3}}{9 x ^{2}} + \frac{1}{9 x ^{2}}

Cancel

\frac{3 x ^{3}}{9 x ^{2}} : \frac{x}{3}

= \frac{x ^{7}}{9} + \frac{x ^{4}}{3} + \frac{x}{3} + \frac{1}{9 x ^{2}}

g (f (x + 1))

- e^{(- 2 x)} - 1

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

(1, - 4, 1) x (2, 3, 0)

(- 17) \times (6)

What is expand ((x^3+1)^3)/(9x^2) ?
The solution to expand ((x^3+1)^3)/(9x^2) is (x^7)/9+(x^4)/3+x/3+1/(9x^2)