What is expand (y(-2y+4)^3)/3 ?

The solution to expand (y(-2y+4)^3)/3 is -(8y^4)/3+16y^3-32y^2+(64y)/3

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expand (y(-2y+4)^3)/3

expand

\frac{y ( - 2 y + 4 ) ^{3}}{3}

- \frac{8 y ^{4}}{3} + 16 y^{3} - 32 y^{2} + \frac{64 y}{3}

Solution steps

\frac{y ( - 2 y + 4 ) ^{3}}{3}

(- 2 y + 4)^{3} = - 8 y^{3} + 48 y^{2} - 96 y + 64

= \frac{y ( - 8 y ^{3} + 48 y ^{2} - 96 y + 64 )}{3}

Expand

y (- 8 y^{3} + 48 y^{2} - 96 y + 64) : - 8 y^{4} + 48 y^{3} - 96 y^{2} + 64 y

= \frac{- 8 y ^{4} + 48 y ^{3} - 96 y ^{2} + 64 y}{3}

Apply the fraction rule:

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

\frac{- 8 y ^{4} + 48 y ^{3} - 96 y ^{2} + 64 y}{3} = - \frac{8 y ^{4}}{3} + \frac{48 y ^{3}}{3} - \frac{96 y ^{2}}{3} + \frac{64 y}{3}

= - \frac{8 y ^{4}}{3} + \frac{48 y ^{3}}{3} - \frac{96 y ^{2}}{3} + \frac{64 y}{3}

Divide the numbers:

\frac{48}{3} = 16

= - \frac{8 y ^{4}}{3} + 16 y^{3} - \frac{96 y ^{2}}{3} + \frac{64 y}{3}

Divide the numbers:

\frac{96}{3} = 32

= - \frac{8 y ^{4}}{3} + 16 y^{3} - 32 y^{2} + \frac{64 y}{3}

\frac{2 x ^{3}}{( 1 + 4 x ) ^{2}}

- 2 (- 5) (3) (- 1)

(- x^{2} + 4 x + 5)^{5}

y (2 x - 3 y) d y

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

What is expand (y(-2y+4)^3)/3 ?
The solution to expand (y(-2y+4)^3)/3 is -(8y^4)/3+16y^3-32y^2+(64y)/3