What is expand (n(n-2)(2n+7))/6 ?

The solution to expand (n(n-2)(2n+7))/6 is (n^3)/3+(n^2)/2-(7n)/3

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expand (n(n-2)(2n+7))/6

expand

\frac{n ( n - 2 ) ( 2 n + 7 )}{6}

\frac{n ^{3}}{3} + \frac{n ^{2}}{2} - \frac{7 n}{3}

Solution steps

\frac{n ( n - 2 ) ( 2 n + 7 )}{6}

Expand

n (n - 2) (2 n + 7) : 2 n^{3} + 3 n^{2} - 14 n

= \frac{2 n ^{3} + 3 n ^{2} - 14 n}{6}

Apply the fraction rule:

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

\frac{2 n ^{3} + 3 n ^{2} - 14 n}{6} = \frac{2 n ^{3}}{6} + \frac{3 n ^{2}}{6} - \frac{14 n}{6}

= \frac{2 n ^{3}}{6} + \frac{3 n ^{2}}{6} - \frac{14 n}{6}

Cancel

\frac{2 n ^{3}}{6} : \frac{n ^{3}}{3}

= \frac{n ^{3}}{3} + \frac{3 n ^{2}}{6} - \frac{14 n}{6}

Cancel

\frac{3 n ^{2}}{6} : \frac{n ^{2}}{2}

= \frac{n ^{3}}{3} + \frac{n ^{2}}{2} - \frac{14 n}{6}

Cancel

\frac{14 n}{6} : \frac{7 n}{3}

= \frac{n ^{3}}{3} + \frac{n ^{2}}{2} - \frac{7 n}{3}

\frac{3 ( 5 - 3 y )}{2}

(- 4 x + 3 y = 2,) (5 x - 4 y = 1,) (3 x + z = 0)

v (6, - 4) f (- 2, - 4)

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

\frac{1 - x ( x - 2 )}{x - 2}

What is expand (n(n-2)(2n+7))/6 ?
The solution to expand (n(n-2)(2n+7))/6 is (n^3)/3+(n^2)/2-(7n)/3