What is expand (5(n+2x-2)^4)/(n^4) ?

The solution to expand (5(n+2x-2)^4)/(n^4) is

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expand (5(n+2x-2)^4)/(n^4)

expand

\frac{5 ( n + 2 x - 2 ) ^{4}}{n ^{4}}

\frac{80 x ^{4}}{n ^{4}} + \frac{160 x ^{3}}{n ^{3}} - \frac{320 x ^{3}}{n ^{4}} + \frac{120 x ^{2}}{n ^{2}} - \frac{480 x ^{2}}{n ^{3}} + \frac{480 x ^{2}}{n ^{4}} + \frac{40 x}{n} - \frac{240 x}{n ^{2}} + \frac{480 x}{n ^{3}} - \frac{320 x}{n ^{4}} + 5 - \frac{40}{n} + \frac{120}{n ^{2}} - \frac{160}{n ^{3}} + \frac{80}{n ^{4}}

Solution steps

\frac{5 ( n + 2 x - 2 ) ^{4}}{n ^{4}}

(n + 2 x - 2)^{4} = 16 x^{4} + 32 n x^{3} - 64 x^{3} + 24 n^{2} x^{2} - 96 n x^{2} + 96 x^{2} + 8 n^{3} x - 48 n^{2} x + 96 n x - 64 x + n^{4} - 8 n^{3} + 24 n^{2} - 32 n + 16

= \frac{5 ( 16 x ^{4} + 32 n x ^{3} - 64 x ^{3} + 24 n ^{2} x ^{2} + 96 x ^{2} - 96 n x ^{2} + 8 n ^{3} x + 96 n x - 48 n ^{2} x - 64 x + n ^{4} + 24 n ^{2} + 16 - 8 n ^{3} - 32 n )}{n ^{4}}

Expand

5 (16 x^{4} + 32 n x^{3} - 64 x^{3} + 24 n^{2} x^{2} - 96 n x^{2} + 96 x^{2} + 8 n^{3} x - 48 n^{2} x + 96 n x - 64 x + n^{4} - 8 n^{3} + 24 n^{2} - 32 n + 16) : 80 x^{4} + 160 n x^{3} - 320 x^{3} + 120 n^{2} x^{2} - 480 n x^{2} + 480 x^{2} + 40 n^{3} x - 240 n^{2} x + 480 n x - 320 x + 5 n^{4} - 40 n^{3} + 120 n^{2} - 160 n + 80

= \frac{80 x ^{4} + 160 n x ^{3} - 320 x ^{3} + 120 n ^{2} x ^{2} - 480 n x ^{2} + 480 x ^{2} + 40 n ^{3} x - 240 n ^{2} x + 480 n x - 320 x + 5 n ^{4} - 40 n ^{3} + 120 n ^{2} - 160 n + 80}{n ^{4}}

Apply the fraction rule:

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

\frac{80 x ^{4} + 160 n x ^{3} - 320 x ^{3} + 120 n ^{2} x ^{2} - 480 n x ^{2} + 480 x ^{2} + 40 n ^{3} x - 240 n ^{2} x + 480 n x - 320 x + 5 n ^{4} - 40 n ^{3} + 120 n ^{2} - 160 n + 80}{n ^{4}} = \frac{80 x ^{4}}{n ^{4}} + \frac{160 n x ^{3}}{n ^{4}} - \frac{320 x ^{3}}{n ^{4}} + \frac{120 n ^{2} x ^{2}}{n ^{4}} - \frac{480 n x ^{2}}{n ^{4}} + \frac{480 x ^{2}}{n ^{4}} + \frac{40 n ^{3} x}{n ^{4}} - \frac{240 n ^{2} x}{n ^{4}} + \frac{480 n x}{n ^{4}} - \frac{320 x}{n ^{4}} + \frac{5 n ^{4}}{n ^{4}} - \frac{40 n ^{3}}{n ^{4}} + \frac{120 n ^{2}}{n ^{4}} - \frac{160 n}{n ^{4}} + \frac{80}{n ^{4}}

= \frac{80 x ^{4}}{n ^{4}} + \frac{160 n x ^{3}}{n ^{4}} - \frac{320 x ^{3}}{n ^{4}} + \frac{120 n ^{2} x ^{2}}{n ^{4}} - \frac{480 n x ^{2}}{n ^{4}} + \frac{480 x ^{2}}{n ^{4}} + \frac{40 n ^{3} x}{n ^{4}} - \frac{240 n ^{2} x}{n ^{4}} + \frac{480 n x}{n ^{4}} - \frac{320 x}{n ^{4}} + \frac{5 n ^{4}}{n ^{4}} - \frac{40 n ^{3}}{n ^{4}} + \frac{120 n ^{2}}{n ^{4}} - \frac{160 n}{n ^{4}} + \frac{80}{n ^{4}}

\frac{160 n x ^{3}}{n ^{4}} = \frac{160 x ^{3}}{n ^{3}}

\frac{120 n ^{2} x ^{2}}{n ^{4}} = \frac{120 x ^{2}}{n ^{2}}

\frac{480 n x ^{2}}{n ^{4}} = \frac{480 x ^{2}}{n ^{3}}

\frac{40 n ^{3} x}{n ^{4}} = \frac{40 x}{n}

\frac{240 n ^{2} x}{n ^{4}} = \frac{240 x}{n ^{2}}

\frac{480 n x}{n ^{4}} = \frac{480 x}{n ^{3}}

\frac{5 n ^{4}}{n ^{4}} = 5

\frac{40 n ^{3}}{n ^{4}} = \frac{40}{n}

\frac{120 n ^{2}}{n ^{4}} = \frac{120}{n ^{2}}

\frac{160 n}{n ^{4}} = \frac{160}{n ^{3}}

= \frac{80 x ^{4}}{n ^{4}} + \frac{160 x ^{3}}{n ^{3}} - \frac{320 x ^{3}}{n ^{4}} + \frac{120 x ^{2}}{n ^{2}} - \frac{480 x ^{2}}{n ^{3}} + \frac{480 x ^{2}}{n ^{4}} + \frac{40 x}{n} - \frac{240 x}{n ^{2}} + \frac{480 x}{n ^{3}} - \frac{320 x}{n ^{4}} + 5 - \frac{40}{n} + \frac{120}{n ^{2}} - \frac{160}{n ^{3}} + \frac{80}{n ^{4}}

1 - x y + x (y - x) \frac{d y}{d x}

((ln (x)) + 1)^{3}

\frac{d}{d t} ((x) t (2 t + 1)^{2})

\frac{10 ( 4 23 - 5 )}{49}

\frac{d}{d v} ((x) (u^{2} v + v))

What is expand (5(n+2x-2)^4)/(n^4) ?
The solution to expand (5(n+2x-2)^4)/(n^4) is