What is expand 8((2n^3+3n^2+n)/(6n^3))-4 ?

The solution to expand 8((2n^3+3n^2+n)/(6n^3))-4 is 8/3+4/n+4/(3n^2)-4

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expand 8((2n^3+3n^2+n)/(6n^3))-4

expand

8 (\frac{2 n ^{3} + 3 n ^{2} + n}{6 n ^{3}}) - 4

\frac{8}{3} + \frac{4}{n} + \frac{4}{3 n ^{2}} - 4

Solution steps

8 (\frac{2 n ^{3} + 3 n ^{2} + n}{6 n ^{3}}) - 4

Remove parentheses:

(a) = a

= 8 \cdot \frac{2 n ^{3} + 3 n ^{2} + n}{6 n ^{3}} - 4

8 \cdot \frac{2 n ^{3} + 3 n ^{2} + n}{6 n ^{3}} = \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}}

= \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}} - 4

Apply the fraction rule:

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

\frac{8 n ^{2} + 12 n + 4}{3 n ^{2}} = \frac{8 n ^{2}}{3 n ^{2}} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}}

= \frac{8 n ^{2}}{3 n ^{2}} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}} - 4

Cancel

\frac{8 n ^{2}}{3 n ^{2}} : \frac{8}{3}

= \frac{8}{3} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}} - 4

Cancel

\frac{12 n}{3 n ^{2}} : \frac{4}{n}

= \frac{8}{3} + \frac{4}{n} + \frac{4}{3 n ^{2}} - 4

2 (x + 3 = 12)

\frac{s x + k}{( x - p ) ^{2}}

\frac{( x + h ) - 5}{( x + h ) ^{2} - 5 ( x + h ) + 6}

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

(2 + 4 i,) (- 6 - 3 i,)

What is expand 8((2n^3+3n^2+n)/(6n^3))-4 ?
The solution to expand 8((2n^3+3n^2+n)/(6n^3))-4 is 8/3+4/n+4/(3n^2)-4