What is expand ((2y-y^2)^3*y)/3 ?

The solution to expand ((2y-y^2)^3*y)/3 is -(y^7)/3+2y^6-4y^5+(8y^4)/3

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expand ((2y-y^2)^3*y)/3

expand

\frac{( 2 y - y ^{2} ) ^{3} \cdot y}{3}

- \frac{y ^{7}}{3} + 2 y^{6} - 4 y^{5} + \frac{8 y ^{4}}{3}

Solution steps

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

(2 y - y^{2})^{3} = 8 y^{3} - 12 y^{4} + 6 y^{5} - y^{6}

= \frac{y ( - y ^{6} + 6 y ^{5} - 12 y ^{4} + 8 y ^{3} )}{3}

Expand

(8 y^{3} - 12 y^{4} + 6 y^{5} - y^{6}) y : 8 y^{4} - 12 y^{5} + 6 y^{6} - y^{7}

= \frac{8 y ^{4} - 12 y ^{5} + 6 y ^{6} - y ^{7}}{3}

Apply the fraction rule:

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

\frac{8 y ^{4} - 12 y ^{5} + 6 y ^{6} - y ^{7}}{3} = \frac{8 y ^{4}}{3} - \frac{12 y ^{5}}{3} + \frac{6 y ^{6}}{3} - \frac{y ^{7}}{3}

= \frac{8 y ^{4}}{3} - \frac{12 y ^{5}}{3} + \frac{6 y ^{6}}{3} - \frac{y ^{7}}{3}

Divide the numbers:

\frac{12}{3} = 4

= \frac{8 y ^{4}}{3} - 4 y^{5} + \frac{6 y ^{6}}{3} - \frac{y ^{7}}{3}

Divide the numbers:

\frac{6}{3} = 2

= \frac{8 y ^{4}}{3} - 4 y^{5} + 2 y^{6} - \frac{y ^{7}}{3}

Rewrite in standard form

= - \frac{y ^{7}}{3} + 2 y^{6} - 4 y^{5} + \frac{8 y ^{4}}{3}

(10 - 6 x) e^{x}

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(121 e^{2 t} sin (t) + 121 e^{2 t} cos (t))^{2}

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What is expand ((2y-y^2)^3*y)/3 ?
The solution to expand ((2y-y^2)^3*y)/3 is -(y^7)/3+2y^6-4y^5+(8y^4)/3