What is expand 1+((3n-3)(3n-4)(3n-5))/2 ?

The solution to expand 1+((3n-3)(3n-4)(3n-5))/2 is (141n)/2+(27n^3)/2-54n^2-29

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expand 1+((3n-3)(3n-4)(3n-5))/2

expand

1 + \frac{( 3 n - 3 ) ( 3 n - 4 ) ( 3 n - 5 )}{2}

\frac{141 n}{2} + \frac{27 n ^{3}}{2} - 54 n^{2} - 29

Solution steps

1 + \frac{( 3 n - 3 ) ( 3 n - 4 ) ( 3 n - 5 )}{2}

Expand

(3 n - 3) (3 n - 4) (3 n - 5) : 27 n^{3} - 108 n^{2} + 141 n - 60

= 1 + \frac{27 n ^{3} - 108 n ^{2} + 141 n - 60}{2}

Apply the fraction rule:

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

\frac{27 n ^{3} - 108 n ^{2} + 141 n - 60}{2} = \frac{27 n ^{3}}{2} - \frac{108 n ^{2}}{2} + \frac{141 n}{2} - \frac{60}{2}

= 1 + \frac{27 n ^{3}}{2} - \frac{108 n ^{2}}{2} + \frac{141 n}{2} - \frac{60}{2}

Divide the numbers:

\frac{108}{2} = 54

= 1 + \frac{27 n ^{3}}{2} - 54 n^{2} + \frac{141 n}{2} - \frac{60}{2}

Divide the numbers:

\frac{60}{2} = 30

= 1 + \frac{27 n ^{3}}{2} - 54 n^{2} + \frac{141 n}{2} - 30

Subtract the numbers:

1 - 30 = - 29

= \frac{141 n}{2} + \frac{27 n ^{3}}{2} - 54 n^{2} - 29

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What is expand 1+((3n-3)(3n-4)(3n-5))/2 ?
The solution to expand 1+((3n-3)(3n-4)(3n-5))/2 is (141n)/2+(27n^3)/2-54n^2-29