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

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

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

expand

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

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

Solution steps

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

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

= \frac{( - x ^{3} - 3 x ^{2} y + 6 x ^{2} - 3 x y ^{2} + 12 x y - 12 x - y ^{3} + 6 y ^{2} - 12 y + 8 )}{3}

Remove parentheses:

(- a) = - a

= \frac{- x ^{3} - 3 x ^{2} y + 6 x ^{2} - 3 x y ^{2} + 12 x y - 12 x - y ^{3} + 6 y ^{2} - 12 y + 8}{3}

Apply the fraction rule:

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

\frac{- x ^{3} - 3 x ^{2} y + 6 x ^{2} - 3 x y ^{2} + 12 x y - 12 x - y ^{3} + 6 y ^{2} - 12 y + 8}{3} = - \frac{x ^{3}}{3} - \frac{3 x ^{2} y}{3} + \frac{6 x ^{2}}{3} - \frac{3 x y ^{2}}{3} + \frac{12 x y}{3} - \frac{12 x}{3} - \frac{y ^{3}}{3} + \frac{6 y ^{2}}{3} - \frac{12 y}{3} + \frac{8}{3}

= - \frac{x ^{3}}{3} - \frac{3 x ^{2} y}{3} + \frac{6 x ^{2}}{3} - \frac{3 x y ^{2}}{3} + \frac{12 x y}{3} - \frac{12 x}{3} - \frac{y ^{3}}{3} + \frac{6 y ^{2}}{3} - \frac{12 y}{3} + \frac{8}{3}

\frac{3 x ^{2} y}{3} = x^{2} y

\frac{6 x ^{2}}{3} = 2 x^{2}

\frac{3 x y ^{2}}{3} = x y^{2}

\frac{12 x y}{3} = 4 x y

\frac{12 x}{3} = 4 x

\frac{6 y ^{2}}{3} = 2 y^{2}

\frac{12 y}{3} = 4 y

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

\frac{1}{( y - 1 ) ^{3}}

z (u, v, w (z, u, v)) (2 u + v - w (z, u, v) + 3 z (u, v, w (z, u, v)) = 0)

126 + 32

- \frac{4 ( - 2 x ^{2} + 5 )}{7 x - 9}

y^{2 d t + (2 y t + 1) d y}

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