What is expand ((1-x-y)^3)/3 ?

The solution to expand ((1-x-y)^3)/3 is -(x^3)/3-x^2y+x^2-xy^2+2xy-x-(y^3)/3+y^2-y+1/3

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expand ((1-x-y)^3)/3

expand

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

- \frac{x ^{3}}{3} - x^{2} y + x^{2} - x y^{2} + 2 x y - x - \frac{y ^{3}}{3} + y^{2} - y + \frac{1}{3}

Solution steps

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

(1 - x - y)^{3} = - x^{3} - 3 x^{2} y + 3 x^{2} - 3 x y^{2} + 6 x y - 3 x - y^{3} + 3 y^{2} - 3 y + 1

= \frac{( - x ^{3} - 3 x ^{2} y + 3 x ^{2} - 3 x y ^{2} + 6 x y - 3 x - y ^{3} + 3 y ^{2} - 3 y + 1 )}{3}

Remove parentheses:

(- a) = - a

= \frac{- x ^{3} - 3 x ^{2} y + 3 x ^{2} - 3 x y ^{2} + 6 x y - 3 x - y ^{3} + 3 y ^{2} - 3 y + 1}{3}

Apply the fraction rule:

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

\frac{- x ^{3} - 3 x ^{2} y + 3 x ^{2} - 3 x y ^{2} + 6 x y - 3 x - y ^{3} + 3 y ^{2} - 3 y + 1}{3} = - \frac{x ^{3}}{3} - \frac{3 x ^{2} y}{3} + \frac{3 x ^{2}}{3} - \frac{3 x y ^{2}}{3} + \frac{6 x y}{3} - \frac{3 x}{3} - \frac{y ^{3}}{3} + \frac{3 y ^{2}}{3} - \frac{3 y}{3} + \frac{1}{3}

= - \frac{x ^{3}}{3} - \frac{3 x ^{2} y}{3} + \frac{3 x ^{2}}{3} - \frac{3 x y ^{2}}{3} + \frac{6 x y}{3} - \frac{3 x}{3} - \frac{y ^{3}}{3} + \frac{3 y ^{2}}{3} - \frac{3 y}{3} + \frac{1}{3}

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

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

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

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

\frac{3 x}{3} = x

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

\frac{3 y}{3} = y

= - \frac{x ^{3}}{3} - x^{2} y + x^{2} - x y^{2} + 2 x y - x - \frac{y ^{3}}{3} + y^{2} - y + \frac{1}{3}

\frac{1}{2 sin ( \frac{1}{2} ( θ + \frac{π}{6} ) ) + 1}

P (3, - 5) Q (- 2, 0)

(1 + e^{x})^{4}

\frac{4 ( x + h ) - 5 - 4 x - 5}{h}

(a + b + c) (d + e)

What is expand ((1-x-y)^3)/3 ?
The solution to expand ((1-x-y)^3)/3 is -(x^3)/3-x^2y+x^2-xy^2+2xy-x-(y^3)/3+y^2-y+1/3