What is expand (-l(-2l(-2l(-2l+1)-1)+1)-1)/8 ?

The solution to expand (-l(-2l(-2l(-2l+1)-1)+1)-1)/8 is l^4-(l^3)/2-(l^2)/4-l/8-1/8

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expand (-l(-2l(-2l(-2l+1)-1)+1)-1)/8

expand

\frac{- l ( - 2 l ( - 2 l ( - 2 l + 1 ) - 1 ) + 1 ) - 1}{8}

l^{4} - \frac{l ^{3}}{2} - \frac{l ^{2}}{4} - \frac{l}{8} - \frac{1}{8}

Solution steps

\frac{- l ( - 2 l ( - 2 l ( - 2 l + 1 ) - 1 ) + 1 ) - 1}{8}

l (- 2 l (- 2 l (- 2 l + 1) - 1) + 1) = l (- 8 l^{3} + 4 l^{2} + 2 l + 1)

= \frac{- l ( - 8 l ^{3} + 4 l ^{2} + 2 l + 1 ) - 1}{8}

Expand

- l (- 8 l^{3} + 4 l^{2} + 2 l + 1) - 1 : 8 l^{4} - 4 l^{3} - 2 l^{2} - l - 1

= \frac{8 l ^{4} - 4 l ^{3} - 2 l ^{2} - l - 1}{8}

Apply the fraction rule:

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

\frac{8 l ^{4} - 4 l ^{3} - 2 l ^{2} - l - 1}{8} = \frac{8 l ^{4}}{8} - \frac{4 l ^{3}}{8} - \frac{2 l ^{2}}{8} - \frac{l}{8} - \frac{1}{8}

= \frac{8 l ^{4}}{8} - \frac{4 l ^{3}}{8} - \frac{2 l ^{2}}{8} - \frac{l}{8} - \frac{1}{8}

Divide the numbers:

\frac{8}{8} = 1

= l^{4} - \frac{4 l ^{3}}{8} - \frac{2 l ^{2}}{8} - \frac{l}{8} - \frac{1}{8}

Cancel

\frac{4 l ^{3}}{8} : \frac{l ^{3}}{2}

= l^{4} - \frac{l ^{3}}{2} - \frac{2 l ^{2}}{8} - \frac{l}{8} - \frac{1}{8}

Cancel

\frac{2 l ^{2}}{8} : \frac{l ^{2}}{4}

= l^{4} - \frac{l ^{3}}{2} - \frac{l ^{2}}{4} - \frac{l}{8} - \frac{1}{8}

\frac{( x - 1 ) lo g _{2} ( x )}{1 + x}

\frac{3 ( 1 + 5 )}{2}

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

\frac{9 ( x ^{10} - 1 )}{5}

\frac{10 x ^{2} + 10}{( x + 1 ) ^{4}}

What is expand (-l(-2l(-2l(-2l+1)-1)+1)-1)/8 ?
The solution to expand (-l(-2l(-2l(-2l+1)-1)+1)-1)/8 is l^4-(l^3)/2-(l^2)/4-l/8-1/8