What is expand ((k-1)(3k^6-k^5-k^4+k^3+k^2+k+1))/(k^4) ?

The solution to expand ((k-1)(3k^6-k^5-k^4+k^3+k^2+k+1))/(k^4) is 3k^3-4k^2+2-1/(k^4)

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expand ((k-1)(3k^6-k^5-k^4+k^3+k^2+k+1))/(k^4)

expand

\frac{( k - 1 ) ( 3 k ^{6} - k ^{5} - k ^{4} + k ^{3} + k ^{2} + k + 1 )}{k ^{4}}

3 k^{3} - 4 k^{2} + 2 - \frac{1}{k ^{4}}

Solution steps

\frac{( k - 1 ) ( 3 k ^{6} - k ^{5} - k ^{4} + k ^{3} + k ^{2} + k + 1 )}{k ^{4}}

Expand

(k - 1) (3 k^{6} - k^{5} - k^{4} + k^{3} + k^{2} + k + 1) : 3 k^{7} - 4 k^{6} + 2 k^{4} - 1

= \frac{3 k ^{7} - 4 k ^{6} + 2 k ^{4} - 1}{k ^{4}}

Apply the fraction rule:

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

\frac{3 k ^{7} - 4 k ^{6} + 2 k ^{4} - 1}{k ^{4}} = \frac{3 k ^{7}}{k ^{4}} - \frac{4 k ^{6}}{k ^{4}} + \frac{2 k ^{4}}{k ^{4}} - \frac{1}{k ^{4}}

= \frac{3 k ^{7}}{k ^{4}} - \frac{4 k ^{6}}{k ^{4}} + \frac{2 k ^{4}}{k ^{4}} - \frac{1}{k ^{4}}

Cancel

\frac{3 k ^{7}}{k ^{4}} : 3 k^{3}

= 3 k^{3} - \frac{4 k ^{6}}{k ^{4}} + \frac{2 k ^{4}}{k ^{4}} - \frac{1}{k ^{4}}

Cancel

\frac{4 k ^{6}}{k ^{4}} : 4 k^{2}

= 3 k^{3} - 4 k^{2} + \frac{2 k ^{4}}{k ^{4}} - \frac{1}{k ^{4}}

Cancel

\frac{2 k ^{4}}{k ^{4}} : 2

= 3 k^{3} - 4 k^{2} + 2 - \frac{1}{k ^{4}}

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What is expand ((k-1)(3k^6-k^5-k^4+k^3+k^2+k+1))/(k^4) ?
The solution to expand ((k-1)(3k^6-k^5-k^4+k^3+k^2+k+1))/(k^4) is 3k^3-4k^2+2-1/(k^4)