What is expand-(9y(y-1)^3)/2 ?

The solution to expand-(9y(y-1)^3)/2 is -(9y^4)/2+(27y^3)/2-(27y^2)/2+(9y)/2

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expand-(9y(y-1)^3)/2

expand

- \frac{9 y ( y - 1 ) ^{3}}{2}

- \frac{9 y ^{4}}{2} + \frac{27 y ^{3}}{2} - \frac{27 y ^{2}}{2} + \frac{9 y}{2}

Solution steps

- \frac{9 y ( y - 1 ) ^{3}}{2}

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

= - \frac{9 y ( y ^{3} - 3 y ^{2} + 3 y - 1 )}{2}

Expand

9 y (y^{3} - 3 y^{2} + 3 y - 1) : 9 y^{4} - 27 y^{3} + 27 y^{2} - 9 y

= - \frac{9 y ^{4} - 27 y ^{3} + 27 y ^{2} - 9 y}{2}

Apply the fraction rule:

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

\frac{9 y ^{4} - 27 y ^{3} + 27 y ^{2} - 9 y}{2} = - (\frac{9 y ^{4}}{2}) - (- \frac{27 y ^{3}}{2}) - (\frac{27 y ^{2}}{2}) - (- \frac{9 y}{2})

= - (\frac{9 y ^{4}}{2}) - (- \frac{27 y ^{3}}{2}) - (\frac{27 y ^{2}}{2}) - (- \frac{9 y}{2})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{9 y ^{4}}{2} + \frac{27 y ^{3}}{2} - \frac{27 y ^{2}}{2} + \frac{9 y}{2}

x (ln (x) + 1) + 2 ln (x)

\frac{5 ( x + h ) ^{2} + 4 ( x + h )}{h}

\frac{9 ( 1 - n ^{3} )}{5}

π (\frac{3}{3} + 1)

(O e + O i) (D i + De)

What is expand-(9y(y-1)^3)/2 ?
The solution to expand-(9y(y-1)^3)/2 is -(9y^4)/2+(27y^3)/2-(27y^2)/2+(9y)/2