What is expand ((-a(n)^3+a(n)*(1+6n)))/6 ?

The solution to expand ((-a(n)^3+a(n)*(1+6n)))/6 is -(an^3)/6+(an)/6+an^2

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expand ((-a(n)^3+a(n)*(1+6n)))/6

expand

\frac{( - a ( n ) ^{3} + a ( n ) \cdot ( 1 + 6 n ) )}{6}

- \frac{a n ^{3}}{6} + \frac{an}{6} + a n^{2}

Solution steps

\frac{- a n ^{3} + a ( n ) ( 1 + 6 n )}{6}

Remove parentheses:

(a) = a

= \frac{- a n ^{3} + an ( 1 + 6 n )}{6}

Expand

- a n^{3} + an (1 + 6 n) : - a n^{3} + an + 6 a n^{2}

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{6}{6} = 1

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

12 (\frac{x - 4}{4} + \frac{x}{3} = \frac{6}{1})

(z + 3)^{2}

\frac{( - x ^{2} - 64 )}{( x ^{2} - 64 ) ^{2}}

\frac{( n + 1 ) ^{2}}{n ^{3}}

What is expand ((-a(n)^3+a(n)*(1+6n)))/6 ?
The solution to expand ((-a(n)^3+a(n)*(1+6n)))/6 is -(an^3)/6+(an)/6+an^2