What is expand k^2((k^2+1)(k^2+2)+k^2+1)/6 ?

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

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expand k^2((k^2+1)(k^2+2)+k^2+1)/6

expand

k^{2} \frac{( k ^{2} + 1 ) ( k ^{2} + 2 ) + k ^{2} + 1}{6}

\frac{k ^{6}}{6} + \frac{2 k ^{4}}{3} + \frac{k ^{2}}{2}

Solution steps

k^{2} \frac{( k ^{2} + 1 ) ( k ^{2} + 2 ) + k ^{2} + 1}{6}

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

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

Multiply fractions:

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

= \frac{( k ^{4} + 4 k ^{2} + 3 ) k ^{2}}{6}

Expand

(k^{4} + 4 k^{2} + 3) k^{2} : k^{6} + 4 k^{4} + 3 k^{2}

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

Apply the fraction rule:

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

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

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

Cancel

\frac{4 k ^{4}}{6} : \frac{2 k ^{4}}{3}

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

Cancel

\frac{3 k ^{2}}{6} : \frac{k ^{2}}{2}

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

\frac{( s ^{4} + 2 s ^{3} + 3 s ^{2} + 4 s + 5 )}{( s ( s + 1 ) )}

\frac{x ^{2} - 6 x - 7}{( x - 3 ) ^{2}}

(x^{3})^{4}

p (2, - 3) Q (3, 5)

Ker (a + b x + c^{2} + d x^{3})

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