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

The solution to expand (y(3-y)^2)/2 is (y^3)/2-3y^2+(9y)/2

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

expand

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

\frac{y ^{3}}{2} - 3 y^{2} + \frac{9 y}{2}

Solution steps

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

(3 - y)^{2} = 9 - 6 y + y^{2}

= \frac{y ( y ^{2} - 6 y + 9 )}{2}

Expand

y (9 - 6 y + y^{2}) : 9 y - 6 y^{2} + y^{3}

= \frac{9 y - 6 y ^{2} + y ^{3}}{2}

Apply the fraction rule:

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

\frac{9 y - 6 y ^{2} + y ^{3}}{2} = \frac{9 y}{2} - \frac{6 y ^{2}}{2} + \frac{y ^{3}}{2}

= \frac{9 y}{2} - \frac{6 y ^{2}}{2} + \frac{y ^{3}}{2}

Divide the numbers:

\frac{6}{2} = 3

= \frac{9 y}{2} - 3 y^{2} + \frac{y ^{3}}{2}

Rewrite in standard form

= \frac{y ^{3}}{2} - 3 y^{2} + \frac{9 y}{2}

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

\frac{6 ( x + h )}{( x + h ) + 2}

\frac{3 ( s + 2 )}{( s + 1 ) ( s + 3 )}

\frac{5 ( x + 2 )}{12}

- 3 (1 + 6 x = 14 + x)

What is expand (y(3-y)^2)/2 ?
The solution to expand (y(3-y)^2)/2 is (y^3)/2-3y^2+(9y)/2