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

The solution to expand ((3x^2-x)^3)/3 is 9x^6-9x^5+3x^4-(x^3)/3

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

expand

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

9 x^{6} - 9 x^{5} + 3 x^{4} - \frac{x ^{3}}{3}

Solution steps

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

(3 x^{2} - x)^{3} = 27 x^{6} - 27 x^{5} + 9 x^{4} - x^{3}

= \frac{27 x ^{6} - 27 x ^{5} + 9 x ^{4} - x ^{3}}{3}

Apply the fraction rule:

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

\frac{27 x ^{6} - 27 x ^{5} + 9 x ^{4} - x ^{3}}{3} = \frac{27 x ^{6}}{3} - \frac{27 x ^{5}}{3} + \frac{9 x ^{4}}{3} - \frac{x ^{3}}{3}

= \frac{27 x ^{6}}{3} - \frac{27 x ^{5}}{3} + \frac{9 x ^{4}}{3} - \frac{x ^{3}}{3}

Divide the numbers:

\frac{27}{3} = 9

= 9 x^{6} - 9 x^{5} + \frac{9 x ^{4}}{3} - \frac{x ^{3}}{3}

Divide the numbers:

\frac{9}{3} = 3

= 9 x^{6} - 9 x^{5} + 3 x^{4} - \frac{x ^{3}}{3}

\frac{1}{s C ( L s + R ) + 1}

\frac{n ( 11 - n )}{2}

\frac{1}{x ( x - 1 ) ( x - 4 )}

ln (x^{2} - 25)

\frac{7}{( x ^{2} - 3 x ) ^{2}}

What is expand ((3x^2-x)^3)/3 ?
The solution to expand ((3x^2-x)^3)/3 is 9x^6-9x^5+3x^4-(x^3)/3