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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

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

Solution steps

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

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

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{12}{3} = 4

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

Divide the numbers:

\frac{6}{3} = 2

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

Rewrite in standard form

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

P (3, - 1) y Q (- 2, - 5)

M (x, y) d x + (x e^{x y} - 6 x y^{2} - 2 y) d y

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

l (- 5 e^{- 2 t} + cos (8 t) + 1)

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

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